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rosijanka [135]
2 years ago
14

An object has a kinetic energy of 810 J after falling a certain distance. If the mass of the object is 20kg, what is the speed o

f the object at this time?
I need the equation and answer please!
Physics
1 answer:
Darina [25.2K]2 years ago
6 0
Don't let the "falling" confuse you. A moving object's kinetic energy is

K.E. = 1/2 (mass) (speed^2)

no matter what direction it's moving, or why.

In this problem, you know the KE and the mass, so

810 J = 1/2 (20 kg) (speed^2).

I think you can probably find the speed now.
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Artist 52 [7]

Answer:

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3 years ago
The acceleration of an object as a function of time is given by a(t) = (1.00 m/s2)t2. If displacement of the object between time
jolli1 [7]

not enough information is given to determine the velocity of the object at time to=0.00s

3 0
3 years ago
How much heat is needed to change the temperature of 3 grams of gold (c = 0.129 ) from 21°C to 363°C? The answer is expressed to
Temka [501]

Q= mcΔT

Where Q is heat or energy

M is mass, c is heat capacitance and t is temperature

You have to convert Celsius into kelvin in order to use this formula I believe

Celsius + 273 = Kelvin

21 + 273 = 294K

363 + 273 = 636K

Now...

Q= (0.003)(0.129)(636-294)

Q= 0.132 J if you are using kilograms, in terms of grams which seems more appropriate the answer would be 132J of energy.  

3 0
2 years ago
Read 2 more answers
A nearsighted eye has a far point of 100 cm. Objects further than 100 cm are not seen clearly. A diverging lens is used to permi
Leya [2.2K]

A diverging lens is used to permit clear vision of an object placed at infinity. The focal length of the lens is -100 cm.

<h3>What is focal length?</h3>

The focal length is half of the radius of curvature of the focal lens.

By the lens maker formula,

1/f = 1/v +1/u

where, v is the image distance and u is the object distance.

Give, the object is at infinity and the image must form at 100 cm, the the focal length will be

1/f = 1/ -100 + 1/∞

f = -100 cm

The focal length must be -100 cm for the diverging lens.

Learn more about focal length.

brainly.com/question/16188698

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6 0
2 years ago
A block of mass m=2.20m=2.20 kg slides down a 30.0^{\circ}30.0
Xelga [282]

Answer:

v_m \approx -4.38\; \rm m \cdot s^{-1} (moving toward the incline.)

v_M \approx 4.02\; \rm m \cdot s^{-1} (moving away from the incline.)

(Assumption: g = 9.81\; \rm m \cdot s^{-2}.)

Explanation:

If g = 9.81\; \rm m \cdot s^{-2}, the potential energy of the block of m = 2.20\; \rm kg would be m \cdot g\cdot h = 2.20\; \rm kg \times 9.81\; \rm m \cdot s^{-2} \times 3.60\; \rm m \approx 77.695\; \rm J when it was at the top of the incline.

If friction is negligible, all these energies would be converted to kinetic energy when this block reaches the bottom of the incline. There shouldn't be any energy loss along the horizontal surface, either. Therefore, the kinetic energy of this m = 2.20\; \rm kg\! block right before the collision would also be approximately 77.695\; \rm J.

Calculate the velocity of that m = 2.20\; \rm kg based on its kinetic energy:

\displaystyle v_m(\text{initial}) = \sqrt{\frac{2\times (\text{Kinetic Energy})}{m}} \approx \sqrt{\frac{2 \times 77.695\; \rm J}{2.20\; \rm kg}} \approx 8.4043\; \rm m \cdot s^{-1}}.

A collision is considered as an elastic collision if both momentum and kinetic energy are conserved.

Initial momentum of the two blocks:

p_m = m \cdot v_m(\text{initial}) \approx 2.20\; \rm kg \times 8.4043\; \rm m \cdot s^{-1} \approx 18.489\; \rm kg \cdot m \cdot s^{-1}.

p_M = M \cdot v_M(\text{initial}) \approx 2.20\; \rm kg \times 0\; \rm m \cdot s^{-1} \approx 0\; \rm kg \cdot m \cdot s^{-1}.

Sum of the momentum of each block right before the collision: approximately 18.489\; \rm kg \cdot m \cdot s^{-1}.

Sum of the momentum of each block right after the collision: (m\cdot v_m + m \cdot v_M).

For momentum to conserve in this collision, v_m and v_M should ensure that m\cdot v_m + m \cdot v_M \approx 18.489\; \rm kg \cdot m \cdot s^{-1}.

Kinetic energy of the two blocks right before the collision: approximately 77.695\; \rm J and 0\; \rm J. Sum of these two values: approximately 77.695\; \rm J\!.

Sum of the energy of each block right after the collision:

\displaystyle \left(\frac{1}{2}\, m \cdot {v_m}^2 + \frac{1}{2}\, M \cdot {v_M}^2\right).

Similarly, for kinetic energy to conserve in this collision, v_m and v_M should ensure that \displaystyle \frac{1}{2}\, m \cdot {v_m}^2 + \frac{1}{2}\, M \cdot {v_M}^2 \approx 77.695\; \rm J.

Combine to obtain two equations about v_m and v_M (given that m = 2.20\; \rm kg whereas M = 7.00\; \rm kg.)

\left\lbrace\begin{aligned}& m\cdot v_m + m \cdot v_M \approx 18.489\; \rm kg \cdot m \cdot s^{-1} \\ & \frac{1}{2}\, m \cdot {v_m}^2 + \frac{1}{2}\, M \cdot {v_M}^2 \approx 77.695\; \rm J\end{aligned}\right..

Solve for v_m and v_M (ignore the root where v_M = 0.)

\left\lbrace\begin{aligned}& v_m \approx -4.38\; \rm m\cdot s^{-1} \\ & v_M \approx 4.02\; \rm m \cdot s^{-1}\end{aligned}\right..

The collision flipped the sign of the velocity of the m = 2.20\; \rm kg block. In other words, this block is moving backwards towards the incline after the collision.

6 0
2 years ago
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