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Dafna11 [192]
3 years ago
5

A wheel of mass M and radius R rolls on a level surface without slipping. If the angular velocity of the wheel about its center

is ?, what is its linear momentum relative to the surface?
Physics
1 answer:
Verizon [17]3 years ago
4 0
The most important thing you should remember in order to get how is its linear momentum relative to the surface is that the<span> point on the circumference is moving at w rad/second. 
Therefore, here is the solving formulae:
if </span>v = R * w you can easily get <span> linear momentum p :
</span> p = M * v = M * R * w
I'm sure that helps.
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A small child has a wagon with a mass of 10 kilograms. The child pulls on the wagon with a force of
Sholpan [36]
F=ma
where:
F - force
m - mass
a - acceleration 

We transform this formula to get a:
a= \frac{F}{m}
a=\frac{2}{10}\frac{N}{kg}=0.2\frac{m}{s^{2}}
4 0
3 years ago
Read 2 more answers
Find the magnitude of the impulse delivered to a soccer ball when a player kicks it with a force of 1360 N. Assume that the play
Alona [7]

Answer:

7.59Ns

Explanation:

Given parameters:

Force  = 1360N

Time of contact  = 5.85 x 10⁻³s

Unknown:

Impulse  = ?

Solution:

The impulse of the ball is given as:

        Impulse  = Force x time

       Impulse  = 1360 x 5.85 x 10⁻³ = 7.59Ns

4 0
2 years ago
If a roller coaster car had 40,000 J of gravitational potential energy when at rest on the top of a hill how much kinetic energy
Inessa [10]

Answer:

K.E = 30,000 J

Explanation:

Given,

The potential energy of the roller coaster car, P.E = 40000 J

The kinetic energy at height h/4, K.E = ?

According to the law of conservation of energy, the total energy of the system is conserved.

At height 'h', the total energy is,

                                    P.E = mgh

                                     K.E = 0

At height 'h/4', the total energy is

                                     P.E + K.E = mgh

                                     P.E = mgh/4

                                     K.E = 1/2 mv²

Therefore,

                                   mgh/4 + 1/2 mv² = mgh

                                    gh/4 + v²/2 = gh

Hence,

                                      v² = 3gh/2

Substituting in the K.E equation

                               K.E = 1/2 mv²

                                      = 1/2 m (3gh/2)

                                       = 3/4 mgh

                                        = 3/4 x 40000

                                         = 30000 J

Hence, the K.E of the roller coaster car is, K.E = 30000 J

6 0
3 years ago
A bomb is dropped from a bomber traveling at the speed of 120 km / h, destroying a military objective located at a distance of 2
schepotkina [342]

Answer:

18 km

Explanation:

Convert km/h to m/s:

120 km/h × (1000 m/km) × (1 h / 3600 s) = 33.3 m/s

The time it takes the bomb to travel the 2000 meters is:

2000 m / (33.3 m/s) = 60 s

So it takes 60 seconds for the bomb to fall.  The distance it fell is therefore:

Δy = v₀ᵧ t + ½ aᵧ t²

Δy = (0 m/s) (60 s) + ½ (10 m/s²) (60 s)²

Δy = 18,000 m

Δy = 18 km

7 0
3 years ago
A locomotive is pulling 8 freight cars, each of which is loaded with the same amount of weight. The mass of each freight car (wi
nikitadnepr [17]

Answer:106560 N

Explanation:

Given

Let Tension between  2 and 3 car be T_{23} and between 3 &4 is T_{34}

T_{23}-T_{34}=ma

mass of freight car =37,000 kg

acceleration of car=0.48 m/s^2

T_{34} is accelerating all freights behind 3

therefore

T_{34}=5\times ma

T_{34}=5\times 37000\times 0.48=88,800 N

Thus

T_{23}=T_{34}+ma

T_{23}=88,800+37000\times 0.48=88,800+17,760=106560 N

4 0
3 years ago
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