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BaLLatris [955]
3 years ago
8

How do you spend a dime and get back a penny

Physics
1 answer:
topjm [15]3 years ago
8 0
Is your daughter doing this is the same time as I do you like you can
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The device used to measure a masses of a body is kilogram . true or false​
olga nikolaevna [1]

Answer: false

Explanation:

While kilograms are the unit used to measure body mass, the device used is a scale.

Hope it helps :)

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
if you can build a cyclotron with twice the radius, by what factor would the allowed maximum particle energy increase? assume th
mina [271]

B. 4

That's the correct answer

6 0
2 years ago
List 6 Physical properties<br><br><br><br><br><br><br> List 6 Chemical properties
nikdorinn [45]
Physical-
Appearance, taste, texture, temperature, state of matter and odor

Chemical-
Reactivity, concentration, Chemical composition, charge, <span>Heat of combustion, and toxicity</span>
3 0
3 years ago
What action involves a chemical change?
Likurg_2 [28]
<span>BAKING A CAKE </span>

Chemical change is a process where a current substance changes or is made into a new type of substance
. Unlike the physical change, which is reversible. Chemical change stays into a its new form. Take for instance these -physical change- examples, making ice cubes. The process involves solidification or freezing where the water becomes ice or solid but when it melts back to its original or typical form with respect to temperature, it’s still water. When the paper is cut into pieces it isn’t burned or exposed to a stimuli that can trigger immediate change in its composition. It’s still the same. On the contrary, baking a cake involves these different compositions or substances –flour, egg, yeast and etc. that is baked to a cake, a newly formed unified substance of all the included ingredients.      



5 0
3 years ago
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