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Mice21 [21]
2 years ago
6

How could you increase the kinetic Energy of an object without changing its mass or gravity? *

Physics
1 answer:
lara [203]2 years ago
8 0

Answer: Height

Explanation:

You could raise it further off the ground.

You might be interested in
Cars A and B are racing each other along the same straight road in the following manner: Car A has a head start and is a distanc
kumpel [21]

Answer:\frac{D_A}{v_B-v_A}

Explanation:

Given

car A had a head start of D_A

and it starts at x=0 and t=0

Car B has to travel a distance of D_A and d_a

where d_a is the distance travel by car A in time t

distance travel by car A is

d_a=v_A\times t

For car B with  speed v_B

d_B=D_A+d_a

v_B\times t=D_A+v_A\times t

t=\frac{D_A}{v_B-v_A}

7 0
3 years ago
50g of ice at 0°C is mixed with 50g of water at 80°C, what will be the final temperature of a mixture in
xxTIMURxx [149]

Answer:

0° C

Explanation:

Given that

Mass of ice, m = 50g

Mass of water, m(w) = 50g

Temperature of ice, T(i) = 0° C

Temperature of water, T(w) = 80° C

Also, it is known that

Specific heat of water, c = 1 cal/g/°C

Latent heat of ice, L(w) = 89 cal/g

Let us assume T to be the final temperature of mixture.

This makes the energy balance equation:

Heat gained by ice to change itself into water + heat gained by melted ice(water) to raise its temperature at T° C = heat lost by water to reach at T° C

m(i).L(i) + m(i).c(w)[T - 0] = m(w).c(w)[80 - T], on substituting, we have

50 * 80 + 50 * 1(T - 0) = 50 * 1(80 - T)

4000 + 50T = 4000 - 50T

0 = 100 T

T = 0° C

Thus, the final temperature is 0° C

3 0
2 years ago
Two charges of equal magnitude Q are held a distance d apart. Consider only points on the line passing through both charges.For
Burka [1]

Answer:

A

Explanation:

6 0
3 years ago
Consider the two moving boxcars in Example 5. Car 1 has a mass of m1 = 65000 kg and a velocity of v01 = +0.80 m/s. Car 2 has a m
Amiraneli [1.4K]

Answer:

1.034m/s

Explanation:

We define the two moments to develop the problem. The first before the collision will be determined by the center of velocity mass, while the second by the momentum preservation. Our values are given by,

m_1 = 65000kg\\v_1 = 0.8m/s\\m_2 = 92000kg\\v_2 = 1.2m/s

<em>Part A)</em> We apply the center of mass for velocity in this case, the equation is given by,

V_{cm} = \frac{m_1v_1+m_2v_2}{m_1+m_2}

Substituting,

V_{cm} = \frac{(65000*0.8)+(92000*1.2)}{92000+65000}

V_{cm} = 1.034m/s

Part B)

For the Part B we need to apply conserving momentum equation, this formula is given by,

m_1v_1+m_2v_2 = (m_1+m_2)v_f

Where here v_f is the velocity after the collision.

v_f = \frac{m_1v_1+m_2v_2}{m_1+m_2}

v_f = \frac{(65000*0.8)+(92000*1.2)}{92000+65000}

v_f = 1.034m/s

8 0
3 years ago
Manipulate p+x=r to solve for x
Tanzania [10]
X=r-p. Maybe I don't understand, but I am assuming that you need to isolate for X? you simply subtract p from both sides.<span />
4 0
2 years ago
Read 2 more answers
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