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nikdorinn [45]
3 years ago
15

A mass M of 3.80E-1 kg slides inside a hoop of radius R=1.10 m with negligible friction. When M is at the top, it has a speed of

5.35 m/s. Calculate the size of the force with which the M pushes on the hoop when M is at an angle of 34.0°.
Physics
1 answer:
Vikentia [17]3 years ago
3 0

Answer:

N = 26.59 N

Explanation:

given,

mass = 0.38 kg

radius of the hoop = 1.10 m

speed = 5.35 m/s

force = ?

now,

\dfrac{1}{2}mv_t^2 + mg(2R) = \dfrac{1}{2}mv^2 + mgR(1-cos \theta)

mv^2 = mv_t^2 + 2mgR(1 + cos \theta)

we know that,

N - mgcos \theta = \dfrac{mv^2}{R}

N - mgcos \theta = \dfrac{mv_t^2 + 2mgR(1 + cos \theta)}{R}

N - mgcos \theta = \dfrac{mv_t^2 }{R}+ 2mg(1 + cos \theta)

N  = \dfrac{mv_t^2 }{R}+ 2mg + 3mgcos \theta)

N  = \dfrac{0.38\times 5.35^2 }{1.1}+ 2\times 0.38\times 9.8 + 3\times 0.38 \times 9.8 cos 34^0)

N = 26.59 N

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Molodets [167]

Answer: I think cars are designed to have crumble zone because lets say you're going 60-70 mph and you hit a brick wall that cant move, it would be a very hard jolt causing the beings inside to get thrown forward, but if it has a crumble zone it would slow the the jolt from is slowing down in the hit.

4 0
3 years ago
In 2007, michael carter (u.s.) set a world record in the shot put with a throw of 24.77 m. what was the initial speed of the sho
Serga [27]

The initial speed of the shot is 15.02 m/s.

The Shot put is released at a height y<em> </em>from the ground with a speed u. It is released at an angle θ to the horizontal. In a time t, the shot put travels a distance <em>R</em> horizontally.

Pl refer to the attached diagram.

Resolve the velocity u into horizontal and vertical components, u ₓ=ucosθ and uy=u sinθ. The horizontal component remains constant in the absence of air resistance, while the vertical component varies due to the action of the gravitational force.

Write an expression for R.

R=u_xt=(ucos \theta)t

Therefore,

t=\frac{R}{ucos\theta} .......(1)

In the time t, the net displacement of the shotput is y in the downward direction.

Use the equation of motion,

y=u_yt-\frac{1}{2}gt^2=(usin\theta) t-\frac{1}{2}gt^2

Substitute the value of t from equation (1).

y=(ucos\theta)(\frac{R}{ucos\theta} )-\frac{1}{2} g(\frac{R}{ucos\theta} )^2\\ =Rtan\theta-(\frac{gR^2}{2u^2cos^2\theta} )

Substitute -2.10 m for y, 24.77 m for R and 38.0° for θ and solve for u.

y=Rtan\theta-(\frac{gR^2}{2u^2cos^2\theta} )\\ (-2.10m)=(24.77 m)(tan38.0^o)-\frac{(9.8 m/s^2)(24.77m)^2}{2u^2(cos38.0^o)^2} \\ u^2=225.71(m/s)^2\\ u=15.02m/s

The shot put was thrown with a speed 15.02 m/s.




7 0
3 years ago
You push a refrigerator with a force of 100 N. If you move the refrigerator a distance of 5 m while you are pushing, how much wo
ddd [48]

The work done to push the refrigerator is 500 Nm.

Explanation:

Work done is the measure of force required to move any object from one point to another. So it is calculated as the product of force and displacement.

If the force increases the work done will increase and similarly, the increase in displacement increases the work done. So to push the refrigerator work should be done on the object and not by the object.

As the force is 100 N and the displacement is 5 m then, work done can be measured as

Work = Force × Displacement

Work = 100 × 5 = 500 Nm

So the work done to push the refrigerator is 500 Nm.

7 0
3 years ago
How long does it take for a truck accelerating at 1.5 m/s^2 to got from rest to 75 km/hr
Free_Kalibri [48]

Answer:

t = 13.9s

Explanation:

u = 0 m/s

v = 75 km/h

= 20.83 m/s

a = 1.5 m/s²

Using

v = u + at

20.83 = 0 + 1.5t

t = 13.9s

6 0
3 years ago
Water is discharged from a pipeline at a velocity v (in ft/sec) given by v = 1306p(1/2), where p is the pressure (in psi). If th
shepuryov [24]

Answer:

a=38.5 ft/sec^{2}

Explanation:

Note that acceleration is the rate change of velocity i.e

acceleration=\frac{change in velocity}{change in time}\\a=\frac{dv}{dt} \\.

Since the velocity is giving as a variable dependent on the pressure, we have to differentiate implicitly both side with respect to time,i.e

\frac{dv}{dt}=1306*(1/2)p^{-1/2}\frac{dp}{dt} \\

if we substitute value for the pressure as giving in the question and also since the rate change of pressure is 0.354psi/sec, we have

a=653*0.1667*0.354\\

a=\frac{dv}{dt}=653(36)^{-1/2}*0.354\\  a=38.5 ft/sec^{2}

3 0
3 years ago
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