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Bess [88]
3 years ago
9

4. A 75 kg bobsled is pushed along a horizontal surface by two athletes. After the

Physics
1 answer:
Alenkinab [10]3 years ago
8 0

The net force on the sled is 300 N

Explanation:

First of all, we start by finding the acceleration of the bobsled, by using the suvat equation:

v^2-u^2=2as

where:

v = 6.0 m/s is the final velocity of the sled

u = 0 is the initial velocity

a is the acceleration

s = 4.5 m is the displacement of the sled

Solving for a, we find

a=\frac{v^2-u^2}{2s}=\frac{6.0^2-0}{2(4.5)}=4 m/s^2

Now we can find the net force on the sled by using Newton's second law:

F = ma

where

F is the net force

m = 75 kg is the mass of the sled

a=4 m/s^2 is the acceleration

Solving the equation, we find the net force:

F=(75)(4)=300 N

Learn more about acceleration and Newton laws here:

brainly.com/question/11411375

brainly.com/question/1971321

brainly.com/question/2286502

brainly.com/question/2562700

#LearnwithBrainly

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A swimming pool is 50 ft wide and 100 ft long and its bottom is an inclined plane, the shallow end having a depth of 4 ft and th
Nina [5.8K]

Explanation:

We define force as the product of mass and acceleration.

F = ma

It means that the object has zero net force when it is in rest state or it when it has no acceleration. However in the case of liquids. just like the above mentioned case, the water is at rest but it is still exerting a pressure on the walls of the swimming pool. That pressure exerted by the liquids in their rest state is known as hydro static force.

Given Data:

Width of the pool = w = 50 ft

length of the pool = l= 100 ft

Depth of the shallow end = h(s) = 4 ft

Depth of the deep end = h(d) = 10 ft.

weight density = ρg = 62.5 lb/ft

Solution:

a) Force on a shallow end:

F = \frac{pgwh}{2} (2x_{1}+h)

F = \frac{(62.5)(50)(4)}{2}(2(0)+4)

F = 25000 lb

b) Force on deep end:

F = \frac{pgwh}{2} (2x_{1}+h)

F = \frac{(62.5)(50)(10)}{2} (2(0)+10)

F = 187500 lb

c) Force on one of the sides:

As it is mentioned in the question that the bottom of the swimming pool is an inclined plane so sum of the forces on the rectangular part and triangular part will give us the force on one of the sides of the pool.

1) Force on the Rectangular part:

F = \frac{pg(l.h)}{2}(2(x_{1} )+ h)

x_{1} = 0\\h_{s} = 4ft

F = \frac{(62.5)(100)(2)}{2}(2(0)+4)

F =25000lb

2) Force on the triangular part:

F = \frac{pg(l.h)}{6} (3x_{1} +2h)

here

h = h(d) - h(s)

h = 10-4

h = 6ft

x_{1} = 4ft\\

F = \frac{62.5 (100)(6)}{6} (3(4)+2(6))

F = 150000 lb

now add both of these forces,

F = 25000lb + 150000lb

F = 175000lb

d) Force on the bottom:

F = \frac{pgw\sqrt{l^{2} + ((h_{d}) - h(s)) } (h_{d}+h_{s})   }{2}

F = \frac{62.5(50)\sqrt{100^{2}(10-4) } (10+4) }{2}

F = 2187937.5 lb

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3 years ago
A car moving at a speed of 36 km/h reaches the foot of a smooth
boyakko [2]

Answer:

d = 10.2 m

Explanation:

When the car travels up the inclined plane, its kinetic energy will be used to do the work in climbing up. So according to the law of conservation of energy, we can write that:

Kinetic\ Energy\ of\ the \ Car = Work\ Done\ while\ moving\ up\ the\ plane\\\frac{1}{2}mv^{2} = Fd

where,

m = mass of car

v = speed of car at the start of plane = (36 km/h)(1000 m/1 km)(1 h/3600 s)

v = 10 m/s

F = force on the car in direction of inclination = W Sin θ

W = weight of car = mg

θ = Angle of inclinition = 30°

d = distance covered up the ramp = ?

Therefore,

\frac{1}{2}mv^{2} = mgdSin\theta\\\frac{1}{2}v^{2} = gdSin\theta\\\frac{1}{2}(10\ m/s)^{2} = d(9.81\ m/s^{2}) Sin\ 30^{0}

<u>d = 10.2 m</u>

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