A 0.38 kg drinking glass is filled with a hot liquid. The liquid transfers 7032 J of energy to the glass. If the
temperature of the glass increases by 22 K, what is the specific heat of the glass?
Answer:
841 J/kg.K
Explanation:
The computation of the specific hear of the glass is shown below:
As we know that
E= cmΔt
where
c denotes specific heat
m denotes 0.38 kg
Δt = temperature = 22k
E denotes energy = 7032 J
Now
7032 J = (0.38) (22) (c)
7032 J = 8.36 (c)
So C = 7032 J ÷ 8.36
= 841 J/kg.K
A steel ball (mass = 50 grams) and a plastic ball of the same dimensions (mass = 10 grams) are dropped from the same height at the same time. Which of the following will occur? (ASSUME NO AIR RESISTANCE)
a. the plastic ball will hit the ground before the steel ball hits the ground
b. the steel ball will hit the ground before the plastic ball hits the ground
c. the steel ball and the plastic ball will hit the ground at the same time
An electric field of intensity 3.25 kN/C is applied along the x-axis. Calculate the electric flux through a rectangular plane 0.350 m wide and 0.700 m long if the following conditions are true. (a) The plane is parallel to the yz-plane. N · m2/C (b) The plane is parallel to the xy-plane. N · m2/C (c) The plane contains the y-axis, and its normal makes an angle of 30.5° with the x-axis. N · m2/C
Answer:
[tex]\varphi_1= 796.25 N m^2/C[/tex]
[tex]\varphi_2= 0 N m^2/C[/tex]
[tex]\varphi_3=686.1 N m^2/C[/tex]
Explanation:
From the question we are told that
Electric field of intensity [tex]E= 3.25 kN/C[/tex]
Rectangle parameter Width [tex]W=0.350 m[/tex] Length [tex]L=0.700 m[/tex]
Angle to the normal [tex]\angle=30.5 \textdegree[/tex]
Generally the equation for Electric flux at parallel to the yz plane [tex]\varphi_1[/tex] is mathematically given by
[tex]\varphi_1=EA cos theta[/tex]
[tex]\varphi_1=3.25* 10^3 N/C * ( 0.350)(0.700) cos 0[/tex]
[tex]\varphi_1= 796.25 N m^2/C[/tex]
Generally the equation for Electric flux at parallel to xy plane [tex]\varphi_2[/tex] is mathematically given by
[tex]\varphi_2=EA cos theta[/tex]
[tex]\varphi_2=3.25* 10^3 N/C * ( 0.350)(0.700) cos 90[/tex]
[tex]\varphi_2= 0 N m^2/C[/tex]
Generally the equation for Electric flux at angle 30 to x plane [tex]\varphi_3[/tex] is mathematically given by
[tex]\varphi_3=EA cos theta[/tex]
[tex]\varphi_3=3.25* 10^3 N/C * ( 0.350)(0.700) cos 30.5[/tex]
[tex]\varphi_3=686.072219 N m^2/C[/tex]
[tex]\varphi_3=686.1 N m^2/C[/tex]
A movie stunt double is supposed to run across the top of a train (in the opposite direction that the train is moving) and just barely jump off before reaching a tunnel, but after reaching the end of the train (starting from the front). If the train is moving at 150 km/hr, is 2 km long and the tunnel is 20 km away from the end (where the stunt double is going to jump from), how fast (in km/hr) will the stunt double need to run
Answer:
the required speed/velocity of the stunt double is 13.633 km/h
Explanation:
Given the data in the question;
velocity of train V = 150 km/h
distance = length of train + distance between the tunnel and the end
= 2 km + 20 km = 22 km
first we calculate time t taken by the train to reach the tunnel;
t = distance / velocity
we substitute
t = 22 km / 250 km/h
t = 0.1467 hr
so the velocity of the of the stunt double will be;
velocity = distance / time
we substitute
velocity = 2 km / 0.1467 hr
velocity = 13.633 km/h
Therefore, the required speed/velocity of the stunt double is 13.633 km/h
A Chinook salmon can jump out of water with a speed of 7.20 m/s . How far horizontally can a Chinook salmon travel through the air if it leaves the water with an initial angle of =29.0° with respect to the horizontal? (Neglect any effects due to air resistance.)
d=
Explanation:
The vertical component of the salmon's velocity is 7.2 m/s x sin 29 = 3.49 m/s
If g = 9.81 m/s^2, the salmon takes
(3.49 m/s) / (9.81 m/s^2) = 0.356 s to reach the highest point of its trajectory.
It takes another 0.356 s to fall back into the water again.
So the salmon is out of the water for a total of 0.712 s.
In this time the salmon travels horizontally with a velocity of 7.2 m/s x cos 29 = 6.30 m/s
We can now calculate the horizontal distance tavelled by multiplying the horizontal velocity by the time spent out of water;
0.712 s x 6.30 m/s = 4.48 m
A ball is rolling across the floor. Why does the ball come to a stop?
The force of gravity stopped it.
The force of friction stopped it.
The normal force stopped it.
It had too much mass.
Explanation:
the notmal focrce stopped it.
Answer:
It's C the friction stopped it
Explanation:
Gravity has a little effect on it. However the main force stopping the ball rolling is the friction force of the floor. The ball will stop rolling when the velocity of the ball is the same as the velocity of friction force.
What is the momentum of a 100-kilogram fullback carrying a football on a play at a velocity of 3.5 m/sec.
Answer:
100 Kg * 3.5 m/sec = 350 Kg-m/s
Explanation:
Momentum= F= Δ(mv) with m= mass, v= velocity, and Δ the change in mass and velocity
In this problem you are given all the factors you need to solve the equation you simply just plug in your mass (1,000 Kg) and Velocity (3.5m/s) and multiply them by each other to get your answer
Radioactive strontium (Sr) is harmful to humans. Strontium, when ingested by cows, can be introduced to milk through the process of replacing an element of the same group or family that has similar properties. Using the Periodic Table, which element do you think can be easily replaced by strontium in milk?
A: B
B: Ag
C: Ca
D: K
Answer:
Ca
Explanation:
It is
A 0.54 kg air hockey puck is initially at rest. What will it's kinetic energy energy be after a net force of 0.56 N acts on it for a distance of 0.84m?
Answer:
Kf = 470 mJ
Explanation:
According the work-energy theorem, the change in the kinetic energy of one object, is equal to the net work done on it.Since the puck is initially at rest, the change is kinetic energy is just the final kinetic energy of the puck.Assuming that the net force is horizontal, and causes a horizontal displacement also, we can find the net work on the puck as follows:[tex]W_{net} = F_{net} * \Delta X = 0.56 N * 0.84 m = 0.47 J = 470 mJ (1)[/tex]
As we have already said, (1) is equal to the final kinetic energy of the puck:⇒ Kf = 470 mJ (2)The membrane that surrounds a certain type of living cell has a surface area of 4.7 x 10-9 m2 and a thickness of 1.3 x 10-8 m. Assume that the membrane behaves like a parallel plate capacitor and has a dielectric constant of 4.7. (a) The potential on the outer surface of the membrane is 79.5 mV greater than that on the inside surface. How much charge resides on the outer surface
Answer:
Q = 1.2*10⁻¹² C
Explanation:
For any capacitor, by definition the capacitance C is equal to the relationship between the charge on one of the conductors and the potential difference between them, as follows:[tex]C = \frac{Q}{V} (1)[/tex]
For the special case of a parallel plate capacitor, just by application of Gauss' law to a rectangular surface half out of the outer surface, and half inside it, it can be showed that the value of the capacitance C is a parameter defined only by geometric constants, as follows:[tex]C = \frac{\epsilon_{0}*\epsilon _{r} * A}{d} (2)[/tex]
So, due to the left sides in (1) and (2) are equal each other, right sides must be equal too.Replacing ε₀, εr (dielectric constant), A, d and V by their values, we can solve for Q, as follows:[tex]Q =\frac{\epsilon_{0} * \epsilon_{r} *A* V}{d} = \frac{(8.85*(4.7)^{2}*79.5)e-24 (F/m*m2*V)}{1.3e-8m} = 1.2e-12 C = 1.2 pC (3)[/tex]
One reason why it’s often easy to miss an action-reaction pair is because of the ________ of one of the objects.
Answer:
an action-reaction pair is because one of the objects is often much more massive and appears to remain motionless when a force acts on it. It has so much inertia, or tendency to remain at rest, that it hardly
How do protons neutrons and electrons differ
Answer/Explanation:
They have a relatively small mass compared to Protons and Neutrons. Protons are electrochemically positive in charge and the Neutrons are electrochemically neutral in charge.
Answer:
They differ because
Proton means positive charge
Electrons are negatively charged
Neutrons are neutral
how do i get the answer for keplers law 3
3.
A student swings a ball attached to the end of a string 0.5m in length in
a vertical circle. The speed of the ball is 2m/s at the highest point and
6m/s at its lowest point: Find the acceleration of the ball at (ii) its highest
point and (ii) its lowest point.
Answer:
I.72m/s²
II.8m/s²
Explanation:
acceleration equal velocity² divided by length
First you lift an object from the floor onto a shelf. Then you move the object back to the floor. do you perform the same amount of work each time? Explain.
What is the power of 600j of work done in 4 seconds?
Explanation:
Power = change in work/change in time
P = 600 joules/ 4 seconds
P= 150 watts
hope this helps :)
The degree of relationship between two or more variables is _________.
a student holds a 3.0kg mass in each hand while sitting on a rotating stool. when his arms are extended horizontally, the masses are 1.0m from the axis of rotation and he rotates with an angular speed of 0.75 rad/s. if the student pulls the masses horizontally to 0.30m from the axis of rotation, what is his new angular speed
Answer:
Explanation:
Due to change in the position of 3 kg mass , the moment of inertia of the system changes , due to which angular speed changes . We shall apply conservation of angular momentum , because no external torque is acting .
Initial moment of inertia I₁ = M R² = 3 x 1 ² = 3 kg m²
Final moment of inertia I₂ = M R² = 3 x .3 ² = 0.27 kg m²
Applying law of conservation of angular momentum
I₁ ω₁ = I₂ ω₂
Putting the values ,
3 x .75 = .27 x ω₂
ω₂ = 8.33 rad / s
New angular speed = 8.33 rad /s .
What is the difference between elastic PE and gravitational PE?
A spider accelerates from a standstill to 5m/s in 10s. What is its acceleration?
Answer:
Acceleration = 0.5m/s²
Explanation:
Given the following data;
Final velocity = 5m/s
Time = 10 seconds
Since the spider started from rest, its initial velocity is equal to 0m/s
To find the acceleration;
In physics, acceleration can be defined as the rate of change of the velocity of an object with respect to time.
This simply means that, acceleration is given by the subtraction of initial velocity from the final velocity all over time.
Mathematically, acceleration is given by the equation;
[tex]Acceleration (a) = \frac{final \; velocity - initial \; velocity}{time}[/tex]
[tex]a = \frac{v - u}{t}[/tex]
Where,
a is acceleration measured in [tex]ms^{-2}[/tex]
v and u is final and initial velocity respectively, measured in [tex]ms^{-1}[/tex]
t is time measured in seconds.
Substituting into the equation, we have;
[tex]a = \frac{5 - 0}{10}[/tex]
[tex]a = \frac{5}{10}[/tex]
Acceleration = 0.5m/s²
The temperature of stars in the universe varies with the type of star and the age of the star among other things. By looking at the shape of the spectrum of light emitted by a star, we can tell something about its average surface temperature.i)If we observe a star's spectrum and find that the peak power occurs at the border between red and infrared light, what is the approximate surface temperature of the star
Answer:
T = 3.6 10³ K
Explanation:
A good way to calculate the surface temperature of a star is to approximate it as a black body, and use Wien's displacement law.
λ T = 2,898 10⁻³
In this case they tell us that the light from the star is between red (700nm) and light infrared (2500nm) suppose that the radiation is
λ = 800 nm (near infrared)
T = 2,898 10⁻³ /λ
T = 2,898 10⁻³/ (800 10⁻⁹)
T = 3.6 10³ K
A police officer is parked by the side of the road, when a speeding car travelling at 50 mi/hrpasses. The police car immediately pursues it, accelerating at a rate of 10 mi/hr per second.The road is fairly busy, so the officer will not go faster than a top speed of 70 mi/hr. How longwill it take before the officer catches up to the speeding car, and how far will it have travelled inorder to do so
Answer:
a) time taken to catch up with speeding car is 12.25 secs
b) the police car will travel 273.8 m to catch up with the speeding car
Explanation:
Given that;
speed of car [tex]V_{c}[/tex] = 50 mi/hr = 22.352 m/s
acceleration of police car = 10 mi/hr = 4.47 m/s²
[tex]V_{f}[/tex] = 70 mi/hr = 31.29 m/s
Now time taken to reach maximum speed is t₁
so
[tex]V_{f}[/tex] = [tex]V_{i}[/tex] + at₁
we substitute
31.29 = 0 + 4.47t₁
t₁ = 31.29 / 4.47
t₁ = 7 sec
now
d₁ = 0 + 1/2 × at₁²
d₁ = 0 + 1/2 × 0 + 4.47×(7)²
d₁ = 109.5 m
so distance travelled by the speeding car in time t₁ will be
[tex]d_{c}[/tex] = [tex]V_{c}[/tex] × t₁
we substitute
[tex]d_{c}[/tex] = 22.352 × 7
[tex]d_{c}[/tex] = 156.46 m
now distance between polive car and speeding car
Δd = [tex]d_{c}[/tex] - d₁
Δd = 156.46 - 109.5
Δd = 46.96 m
time taken to cover Δd will be
t₂ = Δd / ( [tex]V_{f}[/tex] - [tex]V_{c}[/tex] )
t₂ = 46.96 / ( 31.29 - 22.352 )
t₂ = 46.96 / 8.938
t₂ = 5.25 sec
distance travelled by the police in time t₂ will be
d₂ = [tex]V_{f}[/tex] × t₂
d₂ = 31.29 × 5.25
d₂ = 164.3 m
a) How long will it take before the officer catches up to the speeding car;
time taken to catch up with speeding car;
t = t₁ + t₂
t = 7 + 5.25
t = 12.25 secs
Therefore, time taken to catch up with speeding car is 12.25 secs
b) how far will it have travelled in order to do so;
distance = d₁ + d₂
distance = 109.5 + 164.3
distance = 273.8 m
Therefore, the police car will travel 273.8 m to catch up with the speeding car
What is a measure of how much matter an object is made of?
Answer:
grams
Explanation:
How much is the frequency of gold?
Answer:
316 Hz
Explanation:
As for Frequencies of Elements, Gold is said to have 316 Hz
Answer:
The frequency of gold is 316 Hz
Explanation:
Since ancient times, gold is high valued and in the middle ages gold was experienced from the masters of alchemy not only for becoming rich, but also for healing purposes, as from Paracelsus and others.
PLEASE I REALLY NEED HELP!
Question 6
If the car traveled a total of 1,200 meters during this test, what was the average speed of the car? Include the
correct units.
Answer:
[tex]v=\dfrac{1200}{t}\ m/s[/tex]
Explanation:
Given that,
The car traveled a total of 1,200 meters during this test.
We need to find the average speed of the car. The average speed of the car is given by total distance covered divided by the time taken. So,
[tex]v=\dfrac{1200}{t}\ m/s[/tex]
But putting the value of t we can find the average speed of the car.
A 10-meter-long, 150 kg beam extends horizontally from a wall.One end of the beam is fixed to the wall and the other end is attached to the same wall by a cable that makes an angle of 60° with the horizontal. A 75 kg sign is hung from the beam 2.50 meters from the wall.
Determine the magnitude of the tension, in [N] on the cable necessary to keep the system in equilibrium.
Answer:
the magnitude of the tension on the cable necessary to keep the system in equilibrium is 1060.9 N
Explanation:
Given that;
Length L = 10 m
mass of beam m_b = 150 kg; weight W_beam = 150×9.8
mass of sign m = 75 kg
distance of sign hung from the beam from the wall d = 2.50 m
angle ∅ = 60°
g = 9.8 m/s²
Now,
Torque acting at one end of the beam will be;
[tex]T_{net}[/tex] = Tsin∅ × L - mg(d)-W × (L/2)
for equilibrium, [tex]T_{net}[/tex] = 0
therefore, 0 = Tsin∅ × L - mg(d)-W × (L/2)
so we substitute
Tsin(60°) × 10 - 75×9.8(2.50) - 150 × 9.8× (10/2) = 0
Tsin(60°) × 10 - 1837.5 - 7350 = 0
Tsin(60°) × 10 - 9187.5 = 0
Tsin(60°) × 10 = 9187.5
divide both side by 10
Tsin(60°) = 918.75
T × 0.8660 = 918.75
T = 918.75 / 0.8660
T = 1060.9 N
Therefore, the magnitude of the tension on the cable necessary to keep the system in equilibrium is 1060.9 N
The magnitude of the tension on the cable that keep the system in equilibrium is 1060.9 N.
Torque acting at one end of the beam,
= Tsin∅ × L - mg(d)-W × (L/2)
When equilibrium = 0
Tsin∅ × L - mg(d)-W × (L/2) = 0
Where,
L - Length = 10 m
m - mass of sign bord= 75 kg
g- gravitational accelaration = 9.8 m/s²
W - weight of beam = 150×9.8 = 1470 kg
Put the values in the formula,
Tsin(60°) × 10 - 75×9.8(2.50) - 150 × 9.8× (10/2) = 0
Tsin(60°) × 10 - 1837.5 - 7350 = 0
Tsin(60°) × 10 - 9187.5 = 0
Tsin(60°) × 10 = 9187.5
Tsin(60°) = 918.75
T × 0.8660 = 918.75
T = 918.75 / 0.8660
T = 1060.9 N
Therefore, the magnitude of the tension on the cable that keep the system in equilibrium is 1060.9 N.
To know more about magnitude of the tension,
https://brainly.com/question/25403593
what occurred when the photosynthetic began to pump free oxygen into oceans?
when the photosynthetic began to pump free oxygen into oceans, the ocean had enough oxygen to support the life of non-photosynthetic organisms. So, non-photosynthetic organisms came into being.
1. An object that does not give off its own light is called ____
2. The tilt of the Earth's axis creates the
_______
3. Both of the movements of the Earth causes interesting changes on the _______
4. The word solar means of the ______
5. When the Sun, the Earth and the Moon are in line there is an/a _____
Answer:
1. non-luminous objects
sorry have to say the rest in the comments because brai.nly is doing the most for no reason
Explanation:
hope this helps sorry if it doesn't have a good rest of your day/afternoon :) ❤
You serve a volleyball with a mass of 2.1 kg. The ball leaves your hand with a speed of 2.1 m/s. The
ball has ____
energy. Calculate it.
Answer:
4.6 Joules
Explanation:
K=1/2*MV^2
1/2 * 2.1kg * 2.1^2m/s
==4.6305 Joules
simplified to 4.6 Joules
g A boat is anchored 2000 ft from shore anddirects its searchlight towards an automobile travelingdown the straight road. At the particular moment whenthe distance rfrom the searchlight to the automobile is3000 ft, the automobile has speed 80 ft/s and increasesits speed at a rate of 15 ft/s2 down the road.Find the required angular velocity and angularacceleration of the boat's searchlight to track the automobile at this instant.(Answers: 0.018 rad/s CCW, 0.004 rad/s2 CCW)
Answer:
Explanation:
The distance of searchlight will act as radius R and velocity of car may be supposed to be tangential velocity v . We are required to calculate angular velocity ω .
v = 80 ft /s
R = 3000 ft
ω = v / R
= 80 / 3000 = .027 rad / s
For angular acceleration the formula is
angular acceleration α = a / R
a is linear acceleration = 15 ft / s²
α = 15 / 3000 = .005 rad / s².