Answer:
A binary star is a star system consisting of two stars orbiting around their common barycenter.
❣️(◍Jess bregoli◍)❣️#keep learning
A car with a mass of 1 892kg, traveling at 12m/s, was heading towards a wall. It applied 70N of force on its brakes for 6 seconds until it slows down.
Find:
1. The momentum of the car.
2. The resulting impulse of the car.
Answer:
MOMENTUM=205 IMPULSE = 5
Explanation:
The momentum of car is 22704 kg m/s.
The resulting impulse of car is 420 N-s.
What is momentum?Momentum is product of the mass of a particle and its velocity. Momentum is a vector quantity; i.e., it has both magnitude and direction.
p= mv
p= 1892*12
p= 22704 kg m/s
What is impulse?Impulse is the motion produced by a starting force.
[tex]\Delta p = F\Delta t[/tex]
[tex]\Delta p = 70*6\\\Delta p = 420 Ns[/tex]
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The more bonds an atom can make, the more likely it is to combine with other aton
Which element is most likely able to make the greatest variety of bonds?
nitrogen
hydrogen
oxygen
carbon
Answer:
carbon has tetra valency while oxygen can make 2 bonds while nitrogen can make 3 bonds and hydrogen can make 1 bond so i think answer is definatly carbon
Explanation:
calculate the work function that requires a 455 nm photon to eject an electron of a value 0.73 eV
Answer:
W = 2 eV
Explanation:
Given that,
The wavelength of a photon = 455 nm
The kinetic energy of a photon, K = 0.73 eV
We need to find the work function of the electron. It can be solved using Einstein's equation such that,
[tex]W=E-K[/tex]
E is the energy of the photon
So,
[tex]W=\dfrac{hc}{\lambda}-K\\\\W=\dfrac{6.63\times 10^{-34}\times 3\times 10^8}{455\times 10^{-9}\times 1.6\times 10^{-19}}-0.73\\\\W= 2.73\ eV-0.73\ eV\\\\W=2\ eV[/tex]
So, the work function of the metal is 2 eV.
A steel cable lying flat on the floor drags a 20 kg block across a horizontal, frictionless floor. A 100 N force applied to the cable causes the block to reach a speed of 4.2 m/s in a distance of 2.0 m.
What is the mass of the cable?
Answer:
m_cable = 2,676 kg
Explanation:
For this exercise we must look for the acceleration with the kinematic ce relations
v² = v₀² + 2 a x
since the block starts from rest, its initial velocity is vo = 0
a = v² / 2x
a = 4.2² /(2 2.0)
a = 4.41 m / s²
now we can use Newton's second law
Note that the mass that the extreme force has to drag is the mass of the block plus the mass of the cable.
F = (m + m_cable) a
m_cable = F / a -m
m_cable = 100 / 4.41 - 20
m_cable = 2,676 kg
If R1 and R2 are in parallel and R3 is in series with them then equivalent resistance will be
Answer:
Refer to the attachment!~
A 97.6-kg baseball player slides into second base. The coefficient of kinetic friction between the player and the ground is μk = 0.555.
(a) What is the magnitude of the frictional force?
(b) If the player comes to rest after 1.22 s, what is his initial speed?
Answer:
[tex]v=6.65m/sec[/tex]
Explanation:
From the Question we are told that:
Mass [tex]m=97.6[/tex]
Coefficient of kinetic friction [tex]\mu k=0.555[/tex]
Generally the equation for Frictional force is mathematically given by
[tex]F=\mu mg[/tex]
[tex]F=0.555*97.6*9.8[/tex]
[tex]F=531.388N[/tex]
Generally the Newton's equation for Acceleration due to Friction force is mathematically given by
[tex]a_f=-\mu g[/tex]
[tex]a_f=-0.555 *9.81[/tex]
[tex]a_f=-54455m/sec^2[/tex]
Therefore
[tex]v=u-at[/tex]
[tex]v=0+5.45*1.22[/tex]
[tex]v=6.65m/sec[/tex]
A balky cow is leaving the barn as you try harder and harder to push her back in. In coordinates with the origin at the barn door, the cow walks from x = 0 to x = 6.9 m as you apply a force with x-component F _ { x } = - [ 20.0 N + ( 3.0 N / m ) x ]F x =−[20.0N+(3.0N/m)x]. How much work does the force you apply do on the cow during this displacement?
Answer:
The work done is -209.42 J.
Explanation:
F(x) = (- 20 - 3 x ) N
x = 0 to x = 6.9 m
Here, the force is variable in nature, so the work done by the variable force is given by
[tex]W =\int F dx\\\\W =\int_{0}^{6.9}(-20- 3x dx )\\\\W= \left [ - 20 x - 1.5 x^2 \right ]_{0}^{6.9}\\\\W = - 20 (6.9 - 0) - 1.5(6.9\times 6.9 - 0)\\\\W =- 138 - 71.42\\\\W = - 209.42 J[/tex]
Question 1 of 25
Which equation is an example of a synthesis reaction?
A. HNO3 + KOH → KCI + H20
B. 2Li+ CaCl2 - 2LiCl + Ca
O C. S+ 02 - S02
7
O D. CH4 + 202 - 2H2O + CO2
Answer:
C. S + 02 → S02
Explanation:
A synthesis or combination reaction is that reaction involving two elements as reactants to form a single compound as product.
In the reaction given below;
S + 02 → S02
Sulphur and oxygen are elemental substances that combine to synthesize sulfur IV oxide (SO2), and hence it is an example of synthesis reaction.
if a cyclist is travelling a road due east at 12km/h and a wind is blowing from south-west at 5km/h. find the velocity of the wind relative to the cyclist.
Answer:
The velocity of wind with respect to cyclist is [tex]-15.5 \widehat{i} - 3.5 \widehat{j}[/tex].
Explanation:
speed of cyclist = 12 km/h east
speed of wind = 5 km/h south west
Write the speeds in the vector form
[tex]\overrightarrow{vc} = 12 \widehat{i}\\\\overrightarrow{vw} = - 5 (cos 45 \widehat{i} + sin 45 \widehat{j})\\\\\overrightarrow{vw} =-3.5 \widehat{i} - 3.5 \widehat{j}[/tex]
The velocity of wind with respect to cyclist is
[tex]\overrightarrow{vw} =-3.5 \widehat{i} - 3.5 \widehat{j}\overrightarrow{v_(w/c)} = \overrightarrow{vw}-\overrightarrow{vc}\\\\\overrightarrow{v_(w/c)} = - 3.5 \widehat{i} - 3.5 \widehat{j} - 12 \widehat{i}\\\\\overrightarrow{v_(w/c)} =-15.5 \widehat{i} - 3.5 \widehat{j}[/tex]
You sit on ice and shove a heavy box with your feet with a given force. What will you and the box share? *
A) Same acceleration
B) equal and opposite acceleration
C) the equal and opposite force
D) same force
explain please
Answer:
the correct answer is C
Explanation:
In this exercise we will analyze the situation.
When the person is on the ice, the friction coefficient is very small, if the box is in a place where there is no ice, the coefficient is different, so the friction force on each body is different.
Therefore the acceleration that each body acquires is different.
If we apply the conservation of momentum, each body moves in the opposite direction, but with different speeds.
If we use Newton's third law, the force applied to each body has the same magnitude and opposite direction, which is why the force is of the action and reaction type
Consequently the correct answer is C
A gas is enclosed in a cylinder fitted with a light frictionless piston and maintained at atmospheric pressure. When 7950 kcal of heat is added to the gas, the volume is observed to increase slowly from 12.8 m^3 to 19.4 m^3.
Required:
a. Calculate the work done by the gas. Express your answer with the appropriate units.
b. Calculate the change in internal energy of the gas, Express your answer with the appropriate units.
Answer:
Explanation:
From the question we are told that:
Energy [tex]Q=7950kcal=3.3*10^7[/tex]
Initial Volume [tex]V_1=12.8 m^2[/tex]
Final Volume [tex]V_2=19.4 m^2[/tex]
a)
Generally the equation for Work done is mathematically given by
[tex]W=P \triangle V[/tex]
Where
[tex]P= Pressure at 1atm[/tex]
Therefore
[tex]W=(1.01*10^5)(19.4-12.8)[/tex]
[tex]W=6.67*10^5J[/tex]
a)
Generally the equation for Change in internal energy of the gas is mathematically given by
[tex]\triangle U=Q-W[/tex]
[tex]\triangle U=3.3*10^7-6.67*10^5J[/tex]
[tex]\triangle U=3.2*10^7J[/tex]
A bullet is shot straight up into the air from ground level. It reaches a maximum
height at h = 739 m.
Complete question:
A bullet is shot straight up into the air from ground level. It reaches a maximum height at h = 739 m. Calculate the initial velocity of the bullet.
Answer:
the initial velocity of the bullet is 120.35 m/s
Explanation:
Given;
maximum height reached by the bullet, h = 739 m
let the initial velocity of the bullet = u
At maximum height the final velocity of the bullet, v = 0
Apply the following kinematic equation to determine the initial velocity of the bullet.
v² = u² - 2gh
0 = u² - 2gh
u² = 2gh
u = √2gh
u = √(2 x 9.8 x 739)
u = 120.35 m/s
Therefore, the initial velocity of the bullet is 120.35 m/s
Planets closer to a star will have what type of average temperature
Answer:
Mercury - 800°F (430°C) during the day, -290°F (-180°C) at night. Venus - 880°F (471°C) Earth - 61°F (16°C) Mars - minus 20°F (-28°C)30-Jan-2018
New alleles arising from mutations in a population will
A moving object is in equilibrium. What bests describes the motion of the object if mo force changes
•It will change directions
•It will slow down
•It will maintain the state of motion
•It will speed up and ten slow down
Answer:
telekinesis moving things with your mind
a horizontal turntable with radius 0.8 m is rotation about a vertical axis at ts center. a small block sits on the turntable at a distance of 0.4 m from the acis and rotates in a circular path at a speed of 0,800 m/s. what minim coeeficent of static friction between the blok and the turntable is required for the block not to slip
Answer: 0.04
Explanation:
Given
Radius of turntable [tex]r=0.8\ m[/tex]
Block is present at a distance of [tex]r_o=0.4\ m[/tex]
Turntable rotates at a speed of [tex]v=0.8\ m/s[/tex]
angular speed of turntable [tex]\omega =\dfrac{v}{r}[/tex]
[tex]\Rightarrow \omega =\dfrac{0.8}{0.8}\\\Rightarrow \omega =1\ rad/s[/tex]
block will experience a force i.e. centripetal force equal to [tex]m\omega ^2r[/tex]. This must balance the friction force [tex]\mu mg[/tex]
[tex]\Rightarrow m\omega^2 r_o=\mu mg\\\Rightarrow 1^2\times 0.4=\mu \times 9.8\\\Rightarrow \mu =0.04[/tex]
Thus, the coefficient of static friction force is [tex]0.04[/tex]
If a rock displaces 7 ml of water, what is the volume of the rock?
Answer:
if i am not mistaken the volume is 7, because it only took that much space
A 2.2 kg, 20-cm-diameter turntable rotates at 80 rpm on frictionless bearings. Two 600 g blocks fall from above, hit the turntable simultaneously at opposite ends of a diagonal, and stick. What is the turntable's angular velocity, in rpm, just after this event?
Answer:
[tex]w_2=38.3rpm[/tex]
Explanation:
From the question we are told that:
Mass of turntable [tex]M=2.2kg[/tex]
Diameter of turntable [tex]d=20cm=>0.2m[/tex]
Angular Velocity [tex]\omega =80rpm[/tex]
Mass of Blocks [tex]M_b=600g=>0.6kg[/tex]
Generally the equation for inertia is mathematically given by
Initial scenario at \omega=80rpm
[tex]I_1=\frac{1}{2}mR^2[/tex]
[tex]I_1=\frac{1}{2}*2.2*0.1^2[/tex]
[tex]I_1=0.11kgm^2[/tex]
Final scenario
[tex]I_2=I_1+2mR^2[/tex]
[tex]I_2=0.011+(2*0.6*0.12)[/tex]
[tex]I_2=0.023[/tex]
Generally the equation for The relationship between Angular velocity and inertia is mathematically given by
[tex]I_1w_1=I_2w_2[/tex]
[tex]w_2=\frac{I_1 \omega}{I_2}[/tex]
[tex]w_2=\frac{0.011*80}{0.023}[/tex]
[tex]w_2=38.3rpm[/tex]
A speedometer in a car gives the car’s speed at that given moment, or the?
A. General speed
B. Instantaneous speed
C. Average speed
D. Constant speed
It’s not C or D!
Answer:
a because it is at a given moment
Explanation:
did u
Two uncharged metal spheres, spaced 25.0 cm apart, have a capacitance of 26.0 pF. How much work would it take to move 12.0 nC of charge from one sphere to the other? Answer in Joules.
Answer:
[tex]W=2.76\times 10^{-6}\ J[/tex]
Explanation:
Given that,
The distance between two spheres, r = 25 cm = 0.25 m
The capacitance, C = 26 pF = 26×10⁻¹² F
Charge, Q = 12 nC = 12 × 10⁻⁹ C
We need to find the work done in moving the charge. We know that, work done is given by :
[tex]U=\dfrac{Q^2}{2C}[/tex]
Put all the values,
[tex]U=\dfrac{(12\times 10^{-9})^2}{2\times 26\times 10^{-12}}\\\\U=2.76\times 10^{-6}\ J[/tex]
So, the work done is [tex]2.76\times 10^{-6}\ J[/tex].
Imagine that you are standing on a spherical asteroid deep in space far from other objects. You pick up a small rock and throw it straight up from the surface of the asteroid. The asteroid has a radius of 9 m and the rock you threw has a mass of 0.113 kg. You notice that if you throw the rock with a velocity less than 45.7 m/s it eventually comes crashing back into the asteroid.
Required:
Calculate the mass of the asteroid.
Answer:
M = 1.409 10¹⁴ kg
Explanation:
In this exercise we have that the prioress with a minimum speed can escape from the asteroid, therefore we can use the conservation of energy relation.
Starting point. When you drop the stone
Em₀ = K + U
Em₀ = ½ m v² - G m M / r
where M and r are the mass and radius of the asteroid
Final point. When the stone is too far from the asteroid
Em_f = U = - G m M / R_f
as there is no friction, the energy is conserved
Em₀ = Em_f
½ m v² - G m M / r = - G m M / R_f
½ v² = G M (1 / r - 1 /R_f)
indicate that for the speed of v = 45.7 m /s, the stone does not return to the asteroid so R_f = ∞
½ v² = G M (1 /r)
M = [tex]\frac{v^2 r}{2G}[/tex]
let's calculate
M = [tex]\frac{45.7^2 \ 9}{ 2 \ 6.67 \ 10^{-11}}[/tex]
M = 1.409 10¹⁴ kg
How is wind generated?
O A. Air molecules move from areas of low pressure to areas of high
pressure.
O B. Air molecules move more slowly where the temperature is higher
and the pressure is lower.
C. Air molecules move more quickly where the temperature is lower
and the pressure is higher.
D. Air molecules move from areas of high pressure to areas of low
pressure.
Answer:
Explanation:
Wind is caused by the uneven heating of the atmosphere by the sun, variations in the earth's surface, and rotation of the earth. ... Wind turbines convert the energy in wind to electricity by rotating propeller-like blades around a rotor. The rotor turns the drive shaft, which turns an electric generator
A team of people who traveled to the North Pole by dogsled lived on butter because they needed to consume 6 000 dietitian's Calories each day. Because the ice there is lumpy and irregular, they had to help the dogs by pushing and lifting the load. Assume they had a 16-hour working day and that each person could lift a 500-N load. How many times would a person have to lift this weight 1.00 m upwards in a constant gravitational field, where (g = 9.80m/s2) where to do the work equivalent to 6 000 Calories?
Answer:
The right solution is "50200 days".
Explanation:
Given:
Calories intake,
= 6000 kcal,
or,
= [tex]2.52\times 10^7 \ J[/tex]
Force,
= 500 N
As we know,
⇒ [tex]Work \ done = Force\times distance[/tex]
Or,
⇒ [tex]distance = \frac{Work \ done}{Force}[/tex]
By putting the values, we get
[tex]=\frac{2.52\times 10^7}{500}[/tex]
[tex]=0.502\times 10^5[/tex]
[tex]=50200 \ m[/tex]
hence,
The number of days will be:
= [tex]\frac{50200}{1}[/tex]
= [tex]50200 \ days[/tex]
Use a scientific calculator to perform the operation below.
5.92 x 107 + 2.11 x 106
A. 6.13 x 107
B. 2.81 x 101
C. 1.25 x 1014
D. 5.71 x 107
SUBMIT
Answer:
A. 6.13 x 107
Explanation:
Given the expression
5.92 x 10^7 + 2.11 x 10^6
First, we need to convert to the whole number
5.92 x 10^7 = 59200000
2.11 x 10^6 = 2110000
Add both values
59200000 + 2110000
= 61,310,000
Express in standard form
= 6.13 * 10^7
This gives the required result
1. Given an object that follows this time-dependent velocity function: ~v(t) = 2 m/s 2 tˆi − 3 m/s 3 t 2ˆj, and assuming the object begins at the origin at t=0s, where will the object be at t=2.0s?
2. Suppose an object of mass 2.0kg begins at rest and is acted upon by a force F(t) = 5.0N e^−0.1t . What will the object’s speed be after ten seconds?
Answer:
Explanation:
V=ds/dt
Where s =distance traveled
Given
V= 2ti - 3t²j
ds/dt=2ti - 3t²j
ds=(2ti - 3t²j)dt
Integrating
S= 2t²i/2 - 3t³j/3 + C
S=t²i - t³j + C
Since the object starts from Rest when t=0 and s(distance)=0
0=0²i - 0³j + C
C=0
Therefore
S=t²i - t³j
At t=2sec
S=(2)²i - (2)³j
S=4i - 8j
Magnitude of S(distance) = √4²+(-8)²
S= 4√5 meters
S=8.94meters.
2.mass =2kg
F(t)=5e^-0.1t
From Newton 2nd Law
F(t) = mdv/dt
5e^-0.1t= 2dv/dt
2dv = (5e^-0.1t)dt
Integrating
2V(t) = 5e^-0.1t/(-0.1) + C
2v(t) = - 50e^-0.1t + C
Since it started from rest at t=0. That is v(0)=0
2(0) = -50e^-0.1(0) + C
0 = -50 + C
C= 50
v(t) = -50e^-0.1t + 50
At t=10sec
v(10) = -50e^-0.1(10) + 50
V= -18.39 + 50
V= 31.61ms-¹.
While using your calc to evaluate -50e^-0.1(10)
Don't forget the -(minus) sign at the top of the exponential.
If you neglect it... You'll have a different answer.
Hope this helps.
Have a great day!!!
A block whose weight is 45.8 N rests on a horizontal table. A horizontal force of 36.6 N is applied to the block. The coefficients of static and kinetic friction are 0.697 and 0.371, respectively. Will the block move under the influence of the force, and, if so, what will be the block's acceleration? If the block does not move, give 0 m/s2 as the acceleration?
Answer:
Yes it will move and a= 4.19m/s^2
Explanation:
In order for the box to move it needs to overcome the maximum static friction force
Max Static Friction = μFn(normal force)
plug in givens
Max Static friction = 31.9226
Since 36.6>31.9226, the box will move
Mass= Wieght/g which is 45.8/9.8= 4.67kg
Fnet = Fapp-Fk
= 36.6-16.9918
=19.6082
=ma
Solve for a=4.19m/s^2
An athlete swings a ball, connected to the end of a chain, in a horizontal circle. The athlete is able to rotate the ball at the rate of 7.56 rev/s when the length of the chain is 0.600 m. When he increases the length to 0.900 m, he is able to rotate the ball only 6.22 rev/s.
Required:
a. Which rate of rotation gives the greater speed for the ball?
b. What is the centripetal acceleration of the ball at 8.16 rev/s?
c. What is the centripetal acceleration at 6.35 rev/s?
Answer:
a) [tex]\omega _1=7.56rev/s=>47.5rads/s[/tex]
b) [tex]a=30.7[/tex]
c) [tex]a=35.91[/tex]
Explanation:
From the question we are told that:
Initial angular velocity [tex]\omega _1=7.56rev/s=>47.5rads/s[/tex]
Initial Length [tex]L_1=0.600m[/tex]
Final angular velocity [tex]\omega _2=6.22rev/s=39rad/s[/tex]
Final Length [tex]L_2=0.900m[/tex]
a)
Generally the rotation with the greater speed is
[tex]\omega _1=7.56rev/s=>47.5rads/s[/tex]
b)
Generally the equation for centripetal acceleration at 8.16 is mathematically given by
[tex]a=\omega_1^2*L_1[/tex]
[tex]a=8.16 rev/s*0.6[/tex]
[tex]a=30.7[/tex]
c)
At 6.35 rev/s
[tex]a=6.35 rev/s*0.9[/tex]
[tex]a=35.91[/tex]
High speed stroboscopic photographs show that the head of a 183 g golf club is traveling at 58.6 m/s just before it strikes a 46.6 g golf ball at rest on a tee. After the collision, the club head travels (in the same direction) at 40.0 m/s. Find the speed of the golf ball just after impact.
Answer:
The speed of the golf ball just after the impact is 73.04 m/s.
Explanation:
Given that,
The mass of golf club, m₁ = 183 g = 0.183 kg
The mass of golf ball, m₂ = 46.6 g = 0.0466 kg
The initial speed of golf club, u₁ = 58.6 m/s
The initial speed of a golf ball, u₂ = 0
The final speeds of club, v₁ = 40 m/s
We need to find the speed of the golf ball just after impact. Using the conservation of momentum to find it.
[tex]m_1u_1+m_2u_2=m_1v_1+m_2v_2\\\\m_1u_1=m_1v_1+m_2v_2\\\\v_2=\dfrac{m_1u_1-m_1v_1}{m_2}\\\\v_2=\dfrac{0.183 (58.6)-0.183(40)}{0.0466 }\\\\=73.04\ m/s[/tex]
So, the speed of the golf ball just after the impact is 73.04 m/s.
Why do you think we see the sun moves across the sky?
answer it for brainlliest
Answer:
From Earth, the Sun looks like it moves across the sky in the daytime and appears to disappear at night. This is because the Earth is spinning towards the east. The Earth spins about its axis, an imaginary line that runs through the middle of the Earth between the North and South poles.
Nick and Chloe left their campsite by canoe and paddle downstream at an average speed of 12 km/h. They turned around and paddled back upstream at an average speed of 4 km/h. The total trip took 1 hour. After how much time did the campers turn around downstream
The time spent by the campers when they turn around downstream is 15 minutes.
Total distance traveled by Nick and Chloe
The concept of total distance traveled by Nick and Chloe can be used to determine the time they turn around downstream.
Let time for downstream = t1
Let time for upstream = t2
distance covered in upstream = distance covered in downstream = d
12(t1) = d
4(t2) = d
12t1 = 4t2
t1 + t2 = 1
t2 = 1 - t1
12t1 = 4(1 - t1)
12t1 = 4 - 4t1
16t1 = 4
t1 = 4/16
t1 = 0.25 hours
t1 = 0.25(60 min) = 15 mins
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