Answer:
C
Explanation:
Straight away from the source
Answer:c
Explanation:
Straight away from the source
Arctic sea ice has declined over the past few decades causing water levels to increase. This is an interaction of which two spheres? A)Biosphere and geosphere
B)Cryosphere and hydrosphere
C)Geosphere and atmosphere
D)Hydrosphere and geosphere
Answer:
B
Explanation:
Answer:
Its B)Cryosphere and hydrosphere! I took the test and got that question correct!
Explanation:
SOS Physics
Projectile Simulator Lab
Answer:
i will try out give me some time to solve this
5
2. When you keep the mass the same and decrease the unbalanced force how does the acceleration change?
Answer:
The acceleration will decrease due to less force acting on the object.
Explanation:
What effect docs temperature have on the dissolution rate in sugar water?
An astronaut weighing 588 N on earth notices that he weighs only 98 N on moon. His mass on moon is blank kg
Answer:
60 kg
Explanation:
An astronaut weighs 588 N on the earth
He also weighs 98 N on the moon
Therefore the the mass of the astronaut on the moon can be calculated as follows
= 588/98
= 6 × 10
= 60 kg
Hence the mass of the astronaut on the moon is 60 kg
The mass of the astronaut on the moon will be equal to 60 kg whereas the gravitational acceleration on the moon is 1.63 [tex]\rm m/s^2[/tex].
Given information:
An astronaut weighing 588 N on the earth notices that he weighs only 98 N on the moon.
The mass of any object in the whole universe doesn't change but the gravitational force or weight can change. This is because the gravitational acceleration can change.
The mass of the object in the earth or anywhere will be,
[tex]w=mg\\588=m\times 9.8\\m=60\rm \; kg[/tex]
So, the astronaut has 60 kg mass which is constant.
The gravitational acceleration on the moon will be,
[tex]98=60a\\a=1.63\rm\; m/s^2[/tex]
Therefore, the mass of the astronaut on the moon will be equal to 60 kg whereas the gravitational acceleration on the moon is 1.63 [tex]\rm m/s^2[/tex].
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A man is pulling a box with the force of 10N. There is nobody else on the other side of the box. What is the total amount of force on the box? *
Which choice has the least thermal energy?
A swimming pool full of cool water
b all the water in Earth's oceans
C cup of hot cocoa
d bathtub full of warm water
Answer:it’s c a cup of hot cocoa
Explanation:
Answer:
c, cup of hot cocoa
Explanation:
it might seem like the hot cocoa is the hottest, but because the substance is so small, it has the least thermal energy
A 10 kg object has a 40 N force applied to it. What is the acceleration of the object ? *
Answer:
a = 4 [m/s^2]
Explanation:
To solve this problem we must use Newton's second law, which tells us that force is equal to the product of mass by acceleration.
F = m*a
where:
F = force = 40 [N]
m = mass = 10 [kg]
a = 40/10
a = 4 [m/s^2]
mechanical energy defintion
Answer: Mechanical energy is the energy that is possessed by an object due to its motion or due to its position.
(The energy acquired by the objects upon which work is done)
If the forces on an object are balanced, the object will do what?
(a) If two sound waves, one in a gas medium and one in a liquid medium, are equal in intensity, what is the ratio of the pressure amplitude of the wave in the liquid to that of the wave in the gas? Assume that the density of the gas is 2.27 kg/m3 and the density of the liquid is 972 kg/m3. The speed of sound is 376 m/s in the gas medium and 1640 m/s in the liquid. (b) If the pressure amplitudes are equal instead, what is the ratio of the intensities of the waves (of the one in the liquid to that in the gas)?
Answer:
(a) The ratio of the pressure amplitude of the waves is 43.21
(b) The ratio of the intensities of the waves is 0.000535
Explanation:
Given;
density of gas, [tex]\rho _g[/tex] = 2.27 kg/m³
density of liquid, [tex]\rho _l[/tex] = 972 kg/m³
speed of sound in gas, [tex]C_g[/tex] = 376 m/s
speed of sound in liquid, [tex]C_l[/tex] = 1640 m/s
The of the sound wave is given by;
[tex]I = \frac{P_o^2}{2 \rho C} \\\\P_o^2 = 2 \rho C I\\\\p_o = \sqrt{2 \rho CI}[/tex]
Where;
[tex]P_o[/tex] is the pressure amplitude
[tex]P_o_g= \sqrt{2 \rho _g C_gI} -------(1)\\\\P_o_l= \sqrt{2 \rho _l C_lI}---------(2)\\\\\frac{P_o_l}{P_o_g} = \frac{\sqrt{2 \rho _l C_lI}}{\sqrt{2 \rho _g C_gI}} \\\\\frac{P_o_l}{P_o_g} = \sqrt{\frac{2 \rho _l C_lI}{2 \rho _g C_gI} }\\\\ \frac{P_o_l}{P_o_g} = \sqrt{\frac{ \rho _l C_l}{ \rho _g C_g} }\\\\ \frac{P_o_l}{P_o_g} = \sqrt{\frac{ (972)( 1640)}{ (2.27)( 376)} }\\\\\frac{P_o_l}{P_o_g} = 43.21[/tex]
(b) when the pressure amplitudes are equal, the ratio of the intensities is given as;
[tex]I = \frac{P_o^2}{2 \rho C}\\\\I_g = \frac{P_o^2}{2 \rho _g C_g}-------(1)\\\\I_l = \frac{P_o^2}{2 \rho _l C_l}-------(2)\\\\\frac{I_l}{I_g} = (\frac{P_o^2}{2 \rho _l C_l})*(\frac{2\rho_gC_g}{P_o^2} )\\\\\frac{I_l}{I_g} = \frac{\rho _gC_g}{\rho_lC_l} \\\\\frac{I_l}{I_g} = \frac{(2.27)(376)}{(972)(1640)}\\\\ \frac{I_l}{I_g} = 0.000535[/tex]
15. An astronaut on the moon has a 110 kg crate and a 230 kg crate. How do the forces required to liftyihe crates straight up on the moon compare with the forces required to lift them on Earth?
Explanation:
Mass of carte 1 is 110 kg and that of crate 2 is 230 kg
Force required to lift crates is equal to its crate i.e. W = mg
On Moon, a = 1.625 m/s²
Weight of crate 1, W = 110 kg × 1.625 m/s² = 178.75 N
Weight of crate 2, W = 230 kg × 1.625 m/s² = 373.75 N
On Earth, g = 9.8 m/s²
Weight of crate 1, W = 110 kg × 9.8 m/s² = 1078 N
Weight of crate 2, W = 230 kg × 9.8 m/s² = 2254 N
Hence, this is the required solution.
The forces required to lift the crates straight up on the Moon is lesser than the forces required to lift them on Earth.
Given the following data:
Mass A = 110 kgMass B = 230 kgScientific data:
Acceleration due to gravity on Earth = 9.8 [tex]m/s^2[/tex]Acceleration due to gravity on Moon = 1.6 [tex]m/s^2[/tex]The formula for weight.Mathematically, the weight of an object is given by the formula;
[tex]Weight = mg[/tex]
Where;
m is the mass of the object.g is the acceleration due to gravity.For the weight on Moon:[tex]Weight \;A= 110 \times 1.6[/tex]
Weight A = 176 Newton.
[tex]Weight \;B=230 \times 1.6[/tex]
Weight B = 368 Newton.
For the weight on Earth:[tex]Weight \;A= 110 \times 9.8[/tex]
Weight A = 1,078 Newton.
[tex]Weight \;B=230 \times 9.8[/tex]
Weight B = 2,254 Newton.
In conclusion, the forces required to lift the crates straight up on the Moon is lesser than the forces required to lift them on Earth.
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Can someone who knows how to do physics please answer this
Answer:
about 3.17647 hours
Explanation:
The appropriate relation is ...
time = distance/speed
time = (270 km)/(85 km/h) = 3 3/17 h ≈ 3.17647 h
It will take Derek about 3.17647 hours to drive the distance.
a dump truck slowly tilts its bed upward to dispose of a 95 kg steel cabinet. For small angle of tilt the crate stay put,but when the tilt angle exceeds 23.2° the steel cabinet begins to slide.What is the coefficient of static friction between the bed of the truck and the steel cabinet?
Answer:
The value is [tex]\mu =0.4286[/tex]
Explanation:
From the question we are told that
The mass of the steel cabinet is [tex]m = 95 \ kg[/tex]
The threshold angle is [tex]\theta = 23.2^o[/tex]
Generally at the point just before the steel cabinet starts to slide
The horizontal force on the steel cabinet is equal to the frictional force that is
[tex]F_f = F_h[/tex]
Here [tex]F_h[/tex] is the horizontal force which is mathematically represented as
[tex]F_h = m * g cos (theta )[/tex]
and [tex]F_f = \mu N[/tex]
Here N is the normal force acting on the steel cabinet and this is mathematically represented as
[tex]N = mg sin (\theta )[/tex]
=> [tex] m * g cos (theta )= mg sin (\theta ) * \mu [/tex]
=> [tex]\mu = tan (\theta )[/tex]
=> [tex]\mu = tan (23.2)[/tex]
=> [tex]\mu =0.4286[/tex]
The coefficient of static friction between the bed of the truck and steel cabinet is 0.43.
The given parameters;
mass of the steel, m = 95 kgangle of inclination, θ = 23.2⁰The coefficient of static friction between the bed of the truck and steel cabinet is calculated as follows;
[tex]\Sigma F = 0\\\\mg sin(\theta) - \mu mg cos(\theta) = 0\\\\\mu mg cos(\theta ) = mg sin(\theta)\\\\\mu cos(\theta) = sin(\theta)\\\\\mu = \frac{sin(\theta)}{cos(\theta)} \\\\\mu = tan (\theta)\\\\\mu = tan \ (23.2)\\\\\mu = 0.43[/tex]
Thus, the coefficient of static friction between the bed of the truck and steel cabinet is 0.43.
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what is an example of an involuntary muscle?
Answer:
the cardiac muscle is an example
Answer:
your heart
Explanation:
it keeps beating all day and all night
You drive 200 miles in 3 hours before stopping
for 30 minutes for lunch and gas. After lunch,
you travel 150 miles in an hour and a half. What
was your average speed for the trip?
Answer 55.5mph
explanation
is 4x a term? or does it have to remove the 4 to be a term
Answer:
4x is a term oonly
No need to be romoved
HELP PLZ HURRY :((( I NEED IT BY TONIGHT
Answer:
Please see attached image
Explanation:
On his fishing trip Justin takes the boat 4 km south. The fish aren’t biting so he goes 4 km back east. He follows a school of fish 4 km further north and then 3 km west. What distance did he cover? What was his displacement?
Answer:
Distance = 15km
Displacement = 1km
Explanation:
the total distance covered is how far Justin has travelled = 4km + 4km + 4km + 3km
Total distance covered = 15km
To get the displacement, we will use the Pythagoras theorem. Find the diagram attached.
From the diagram, the displacement D is expressed as:
D = AC - BC
From the diagram
AC = 4km
BC = 3km
D = 4km - 3km
D = 1km
Hence his displacement is 1km
HELP PLSSSSSSSSSjjdndnsnsj
Answer:
i feel like 3 not too sure tho
An airplane is traveling at 250 m/s in level flight. If the airplane is to make a change in direction, it must travel is a horizontal curved path. To fly in the curved path, the pilot banks the airplane at an angle such that the lift has a horizontal component that provides the horizontal centripetal acceleration to move in a horizontal circular path. If the airplane is banked at an angle of 15.0 degrees, then the radius of curvature of the curved path of the airplane is
Answer:
The radius of curvature of the curved path of the airplane is 23784.356 meters (23.784 kilometers).
Explanation:
We assume that airplane can be represented as a particle. The free body diagram of the vehicle is presented below as attachment, whose variables are:
[tex]W[/tex] - Weight of the airplane, measured in newtons.
[tex]F[/tex] - Lift, measured in newtons.
[tex]\theta[/tex] - Banking angle, measured in sexagesimal degrees.
The equations of equilibrium associated with the airplane are, respectively:
[tex]\Sigma F_{r} = F\cdot \sin \theta = m\cdot \frac{v^{2}}{R}[/tex] (Eq. 1)
[tex]\Sigma F_{z} = F\cdot \cos \theta - W = 0[/tex] (Eq. 2)
From (Eq. 2):
[tex]F = \frac{W}{\cos \theta}[/tex]
In (Eq. 1):
[tex]W\cdot \tan \theta = m\cdot \frac{v^{2}}{R}[/tex]
By using the definition of weight, we eliminate the mass of the airplane:
[tex]g\cdot \tan \theta = \frac{v^{2}}{R}[/tex]
Where:
[tex]g[/tex] - Gravitational acceleration, measured in meters per square second.
[tex]v[/tex] - Speed, measured in meters per second.
[tex]R[/tex] - Radius of curvature, measured in meters.
Lastly, we clear the radius of curvature with the expression:
[tex]R = \frac{v^{2}}{g\cdot \tan \theta}[/tex]
If we know that [tex]v = 250\,\frac{m}{s}[/tex], [tex]g = 9.807\,\frac{m}{s^{2}}[/tex] and [tex]\theta = 15^{\circ}[/tex], the radius of curvature is:
[tex]R = \frac{\left(250\,\frac{m}{s} \right)^{2}}{\left(9.807\,\frac{m}{s^{2}} \right)\cdot \tan 15^{\circ}}[/tex]
[tex]R = 23784.356\,m[/tex]
The radius of curvature of the curved path of the airplane is 23784.356 meters (23.784 kilometers).
A motorcycle is stopped at a stop light. When the light turns green it
accelerates at 4.2 m/s^2. How far does it travel during 3 s? *
Answer: 18.9 m
Explanation:
i did the kinematic equation & found the answer.
a blackbody is at temperature of 500°C . What would be its temperature in Kelvin for it to radiate twice as much energy per second?
Answer:
Its temperature in Kelvin for it to radiate twice as much energy = [tex]919.257k[/tex]Explanation:
Energy = [tex]6AT^4[/tex]
therefore,
[tex]\frac{E_1}{E_2} = \frac{T_1^4}{T_2^4}[/tex]
where [tex]E_2 = 2E_1[/tex]
therefore,
[tex]\frac{E_1}{2E_1} = \frac{(500+273)^4}{T_2^4}\\\\T_2 = 919.257k\\\\T_2 = 646.257^oC[/tex]
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I WILL MARK YOU AS BRAINLIEST IF RIGHT
A 1125 kg car was traveling at 20 m/s toward a traffic signal when the light suddenly turned red. The driver stomped on her accelerator, clearing the intersection in 1.7 seconds and coming to a speed of 24.6 m/s . What was the acceleration of the car?
Answer:
a = 3.88 m/s^2 (3.88 meters per second squared)
Explanation:
The acceleration of the car can be obtained based of the change of velocities (from initial velocity "vi" to final velocity "vf") in the given amount of time:
In our case:
a = (26.6 - 20) / 1.7 = 3.88 m/s^2
What was the instantaneous speed of Bear A at 5 seconds in meters/min? (No unit conversions necessary, no units required for answer. Type your answer.)
Answer: The answer is 40.
Explanation: There is a typo, they mean 5 minutes. Five minutes is the line between zero and 10, and the dot matches up at 5 and 200. 200 divided by 5= 40.
If an object starts out at rest and accelerates to 100 m/s what is its initial speed?
Answer:0m/s
Explanation.
The initial velocity is 0 because when the object starts moving its speed is zero
A hiker walks 4.30 km east, turns 90.0º due south and walks an additional 2.48 km. What distance did the hiker travel? *
24.64 km
6.78 km
1.82 km
4.96 km
Answer:
6.78 km
Explanation:
Length of path due east = 4.3km
Length of path south = 2.48km
Unknown:
Distance covered = ?
Solution:
The distance covered is the total length of path from start to finish. It takes cognizance of the turns and every direction moved.
Unlike displacement which only considers the net direction from start to finish, distance sums up the total path.
So;
Distance = 4.3km + 2.48km = 6.78km
URGENT!! A 0.057 kg tennis ball and a tennis racket collide. The racket has an initial
momentum of -2.80 kg m/s and a final momentum of -1.97 kg-m/s. The ball
has an initial momentum of 0.02565 kg-m/s. If you assume the collision is
elastic, what is the final velocity of the ball?
O A. -83.26 m/s
O B. -0.80 m/s
O C. -14.11 m/s
O D. -4.75 m/s
Answer:
-14.11
Explanation:
A freight train car has a mass of 2,000 kilograms and an acceleration of 1.8 m/s/s. What is the average force behind that train car?
Answer:
F = 3600 [N]
Explanation:
To solve this problem we must use Newton's second law, which tells us that the sum of force must be equal to the product of mass by acceleration.
ΣF = m*a
where:
F = force [N]
m = mass = 2000 [kg]
a = acceleration = 1.8 [m/s^2]
Now replacing:
F = 2000*1.8
F = 3600 [N]
A disc with moment of inertia 1.5 kg m2 is rotating with an angular speed of 800 rev/min on a shaft. A second disc, initially at rest and with moment of inertia 5.0 kg m2,is suddenly clamped together to the same shaft.
(a) Determine magnitude of angular momentum of the first disc.
(b) Determine the common angular speed of the combination of the two discs.
Answer:
Explanation:
Angular momentum p = Iw
I is the moment of inertia
w is the angular velocity
a) Angular momentum of the first disc
p1 = I1w1
Given
I1 = 1.5kg/m²
w1 = 800rev/min
Convert to rad/s
1 rev = 6.283rad
1 min = 60secs
800rev/min = 800(6.283)/60
= 83.77rad/s
p1 = 1.5×83.77
p1 = 125.66kgrad/s
Hence the magnitude of angular momentum of the first disc is 125.66kgrad/s
b) according to law of conservation of momentum
I1w1+I2w2 = (I1+I2)w
w is the common angular speed
Given
11w1 = 125.66
w2 = 0rad/s
I1 = 1.5kgm²
I2 = 5kgm²
Substitute
125.66+0 = (1.5+5)w
125.66 = 6.6w
w = 125.66/6.5
w = 19.33rad/s
Since 1rad/s = 9.549tev/min
19.33rad/s = 184.6rev/min
Hence the common angular speed of the combination of the two discs is 184.6rev/min