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nikdorinn [45]
3 years ago
10

HELP MEEEEE HURRRYYYY EASYY

Physics
1 answer:
alisha [4.7K]3 years ago
7 0

Answer:

C

Explanation:

thats what i think it depends on the amount of force someone dropped the ball with.

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A .5 kg toy train car moving forward at 3 m/s collides with and sticks to a .8 kg toy car that is traveling at 2 m/s what is the
Viktor [21]
Here we have perfectly inelastic collision. Perfectly inelastic collision is type of collision during which two objects collide, stay connected and momentum is conserved. Formula used for conservation of momentum is:
m_{1} * v_{1} + m_{2} * v_{2} = m_{1} * v'_{1}+ m_{2} * v'_{2}

In case of perfectly inelastic collision v'1 and v'2 are same.

We are given information:
m₁=0.5kg
m₂=0.8kg
v₁=3m/s
v₂=2m/s
v'₁=v'₂=x

0.5*3 + 0.8*2 = 0.5*x + 0.8*x
1.5 + 1.6 = 1.3x
3.1 = 1.3x
x = 2.4 m/s
4 0
2 years ago
A student, starting from rest, slides down a water slide. On the way down, a kinetic frictional force (a nonconservative force)
salantis [7]

Answer:

<em>The velocity with which the student goes down the bottom of glide is 12.48m/s.</em>

Explanation:

The Non conservative force is defined as a force which do not store energy or get he energy dissipate the energy from the system as the system progress with the motion.

Given are

   <em>  mass of the student 73 kg</em>

<em>      height of water glide 11.8 m</em>

<em>      work done as -5.5*10³ J</em>

Have to find speed at which the student goes down the glide.

According to<em> Law of Conservation of energy</em>,

          K.E =P.E+Work Done

 mv²/2=mgh +W

Rearranging the above eqn for v

v = √2(gh+W/m)

Substituting values,

V =  12.48 m/s.

<em>The velocity with which the student goes down the bottom of glide is 12.48m/s.</em>

 

3 0
3 years ago
A block has a volume of 0.09 m3 and a density of 4,000 kg/m3. What's the force of gravity acting on the block in water?
12345 [234]

                                       Density = (mass) / (volume)

                                4,000 kg/m³ = (mass) / (0.09 m³)

Multiply each side
by  0.09 m³ :           (4,000 kg/m³) x (0.09 m³) = mass

                                 mass = 360 kg .

Force of gravity = (mass) x (acceleration of gravity)

                           = (360 kg) x (9.8 m/s²)

                           = (360 x 9.8)  kg-m/s²

                           =   3,528 newtons . 

That's the force of gravity on this block, and it doesn't matter
what else is around it.  It could be in a box on the shelf or at
the bottom of a swimming pool . . . it's weight is 3,528 newtons
(about 793.7 pounds).

Now, it won't seem that heavy when it's in the water, because
there's another force acting on it in the upward direction, against
gravity.  That's the buoyant force due to the displaced water.

The block is displacing 0.09 m³ of water.  Water has 1,000 kg of
mass in a m³, so the block displaces 90 kg of water.  The weight
of that water is  (90) x (9.8) = 882 newtons (about 198.4 pounds),
and that force tries to hold the block up, against gravity.

So while it's in the water, the block seems to weigh

       (3,528  -  882) = 2,646 newtons  (about 595.2 pounds) .

But again ... it's not correct to call that the "force of gravity acting
on the block in water".  The force of gravity doesn't change, but
there's another force, working against gravity, in the water.
5 0
3 years ago
Read 2 more answers
An object was thrown at a certain angle above the ground.It reaches a maximum height of 42.50 meters and hits back the ground 76
aksik [14]

Answer:

Explanation:  Please see my attached calculations.

7 0
3 years ago
A 35 kg box rests on the back of a truck. The coefficient of static friction bet?005 (part 1 of 2)A 35 kg box rests on the back
elena-14-01-66 [18.8K]

Answer with Explanation:

We are given that

Mass of box=35 kg

Coefficient of static friction between box and truck bed=0.202

Acceleration due to gravity=9.8 m/s^2

a.We have to find the force by which the box accelerates forward.

Force by which box accelerates=\mu mg=0.202\times 9.8\times 35

Force by which box accelerates=62.286 N

b.We have to find the maximum acceleration can the truck have before the box slides.

Force =friction force

ma=\mu mg

a=\mu g=9.8\times 0.202=1.9796 m/s^2

Hence, the truck can have maximum acceleration before the box slide=1.9796 m/s^2

3 0
3 years ago
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