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vodomira [7]
3 years ago
12

A 10.00-kilogram block slides along a horizontal, frictionless surface at 12.0 meters per second for 6.00 seconds. The magnitude

of the block's momentum is
Physics
1 answer:
earnstyle [38]3 years ago
4 0

Answer:

120 kg m/s

Explanation:

The magnitude of the momentum of an object is given by

p=mv

where

m is the mass of the object

v is its speed

For the block in this problem,

m = 10.0 kg (mass of the block)

v = 12.0 m/s (speed of the block)

Therefore the magnitude of the block's momentum is

p=(10.0 kg)(12.0 m/s)=120 kg m/s

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a ball is thrown inclined in to the air with a initial velocity of u. if it reaches the maximum height in 6 seconds , find the r
olga55 [171]

Answer:

11:1

Explanation:

At constant acceleration, an object's position is:

y = y₀ + v₀ t + ½ at²

Given y₀ = 0, v₀ = u, and a = -g:

y = u t − ½g t²

After 6 seconds, the ball reaches the maximum height (v = 0).

v = at + v₀

0 = (-g)(6) + u

u = 6g

Substituting:

y = 6g t − ½g t²

The displacement between t=0 and t=1 is:

Δy = [ 6g (1) − ½g (1)² ] − [ 6g (0) − ½g (0)² ]

Δy = 6g − ½g

Δy = 5½g

The displacement between t=6 and t=7 is:

Δy = [ 6g (7) − ½g (7)² ] − [ 6g (6) − ½g (6)² ]

Δy = (42g − 24½g) − (36g − 18g)

Δy = 17½g − 18g

Δy = -½g

So the ratio of the distances traveled is:

(5½g) / (½g)

11 / 1

The ratio is 11:1.

8 0
3 years ago
A student is transmitting sound waves through various materials. Through which metal in the table will the sound waves travel th
MariettaO [177]

Answer:

Aluminum

Explanation:

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5 0
2 years ago
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What would be the result of an alpha particle coming into a magnetic field?
Aneli [31]
The right answer for the question that is being asked and shown above is that: "C) The alpha particle will be deflected in a curve path. " the result of an alpha particle coming into a magnetic field is that <span>C) The alpha particle will be deflected in a curve path. </span>
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3 years ago
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Which SI units are combined to describe a force
Keith_Richards [23]

Answer:

C.

Explanation:

kilograms and m/s^2

4 0
2 years ago
A 79.7 kg base runner begins his slide into second base while moving at a speed of 4.77 m/s. The coefficient of friction between
SashulF [63]

To solve this problem we will apply the concept related to the kinetic energy theorem. Said theorem states that the work done by the net force (sum of all forces) applied to a particle is equal to the change experienced by the kinetic energy of that particle. This is:

\Delta W = \Delta KE

\Delta W = \frac{1}{2} mv^2

Here,

m = mass

v = Velocity

Our values are given as,

m = 79.7kg

v = 4.77m/s

Replacing,

\Delta W = \frac{1}{2} (79.7kg)(4.77m/s)^2

\Delta W = 907J

Therefore the mechanical energy lost due to friction acting on the runner is 907J

6 0
3 years ago
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