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natulia [17]
3 years ago
6

A 7.00- kg bowling ball moves at 3.00 m/s. How fast musta

Physics
1 answer:
ratelena [41]3 years ago
6 0

Answer:

Ping Pong ball move at velocity 160.4 m/s

Explanation:

given data

mass m1 = 7 kg

velocity v1 = 3 m/s

mas m2 = 2.45 g = 2.45 × 10^{-3} kg

same kinetic energy

to find out

How fast Ping Pong ball move (v2)

solution

we know same KE

so

\frac{1}{2}* m1* v1^{2} = \frac{1}{2}* m2* v2^{2}   ...........1

so v2 will be

v2 = \sqrt{\frac{m1*v1^2}{m2} }    .............2

put here value in equation 2 we get v2

v2 = \sqrt{\frac{7*3^2}{2.45*10^{-3}}}

v2 = 160.4

so Ping Pong ball move at velocity 160.4 m/s

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violet has the highest frequency

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3 years ago
What is the effect in the picture
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The effect of this problem is that negative particles and positive particles contract to each other caused by electrical force.
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3 years ago
Jerry the mouse is running along a straight desert road at a constant velocity of 18 m/s. If a certain Tom cat wants to capture
Kruka [31]

Answer:

a) t = 1.75 s

b) x = 31.5 m

Explanation:

a) The time at which Tom should drop the net can be found using the following equation:

y_{f} = y_{0} + v_{oy}t - \frac{1}{2}gt^{2}

Where:

y_{f}: is the final height = 0

y₀: is the initial height = 15 m

g: is the gravity = 9.81 m/s²

v_{0y}: is the initial vertical velocity of the net = 0 (it is dropped from rest)

0 = 15m - \frac{1}{2}9.81 m/s^{2}*t^{2}

t = \sqrt{\frac{2*15 m}{9.81 m/s^{2}}} = 1.75 s

Hence, Tom should drop the net at 1.75 s before Jerry is under the bridge.

b) We can find the distance at which is Jerry when Tom drops the net as follows:

v = \frac{x}{t}

x = v*t = 18 m/s*1.75 m = 31.5 m

Then, Jerry is at 31.5 meters from the bridge when Jerry drops the net.

I hope it helps you!                                                                    

3 0
3 years ago
In a pilot study to help train a group of data collectors who will evaluate childrens pain using a FACES rating scale, the resea
muminat

Answer:

Option c. Inter-rater Reliability

Explanation:

Here, the rating is done by a group of data collectors under training for evaluation children's pain on Faces scale which is a scale ranging from 0 to 10 with different expressions or faces with a happy face at 0 to a crying face at 10.

Also in Inter-rater Reliability, the relative consistency of a study or test is assessed and the extent to which different group members rated the same behavior, the consistency of which is evaluated.

Thus it can also be helpful in interviews, etc.

6 0
4 years ago
A block weighing 400 kg rests on a horizontal surface and supports on top of it ,another block of weight 100 kg which is attache
Paladinen [302]

Answer:

F_a=1470\ N

Explanation:

<u>Friction Force</u>

When objects are in contact with other objects or rough surfaces, the friction forces appear when we try to move them with respect to each other. The friction forces always have a direction opposite to the intended motion, i.e. if the object is pushed to the right, the friction force is exerted to the left.

There are two blocks, one of 400 kg on a horizontal surface and other of 100 kg on top of it tied to a vertical wall by a string. If we try to push the first block, it will not move freely, because two friction forces appear: one exerted by the surface and the other exerted by the contact between both blocks. Let's call them Fr1 and Fr2 respectively. The block 2 is attached to the wall by a string, so it won't simply move with the block 1.  

Please find the free body diagrams in the figure provided below.

The equilibrium condition for the mass 1 is

\displaystyle F_a-F_{r1}-F_{r2}=m.a=0

The mass m1 is being pushed by the force Fa so that slipping with the mass m2 barely occurs, thus the system is not moving, and a=0. Solving for Fa

\displaystyle F_a=F_{r1}+F_{r2}.....[1]

The mass 2 is tried to be pushed to the right by the friction force Fr2 between them, but the string keeps it fixed in position with the tension T. The equation in the horizontal axis is

\displaystyle F_{r2}-T=0

The friction forces are computed by

\displaystyle F_{r2}=\mu \ N_2=\mu\ m_2\ g

\displaystyle F_{r1}=\mu \ N_1=\mu(m_1+m_2)g

Recall N1 is the reaction of the surface on mass m1 which holds a total mass of m1+m2.

Replacing in [1]

\displaystyle F_{a}=\mu \ m_2\ g\ +\mu(m_1+m_2)g

Simplifying

\displaystyle F_{a}=\mu \ g(m_1+2\ m_2)

Plugging in the values

\displaystyle F_{a}=0.25(9.8)[400+2(100)]

\boxed{F_a=1470\ N}

8 0
3 years ago
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