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lana [24]
3 years ago
9

World-class sprinters can accelerate out of the starting blocks with an acceleration that is nearly horizontal and has magnitude

15m/s^2 .
Part A
How much horizontal force F must a sprinter of mass 64 kg exert on the starting blocks to produce this acceleration?
Express your answer in newtons using two significant figures.

Part B
Which body exerts the force that propels the sprinter, the blocks or the sprinter?
Which body exerts the force that propels the sprinter, the blocks or the sprinter?
Physics
1 answer:
Karo-lina-s [1.5K]3 years ago
8 0

Answer:

A. The force exerted by the sprinter must be  9.6 × 10² N.

B. The force that propels the sprinter is exerted by the blocks.

Explanation:

Hi there!

Let´s begin with part B:

The sprinter exerts a force on the blocks and, as a reaction, the blocks exert a force on the sprinter that is of equal magnitude but opposite direction (Newton´s third law). This reaction of the blocks causes the acceleration of the sprinter.

Part A

The force exerted by the blocks can be calculated using Newton´s second law:

F = m · a

Where:

F = exerted force.

m = mass of the object being accelerated.

a = acceleration of the object after applying the force on the object.

F = m · a

F = 64 kg · 15 m/s²

F = 9.6 × 10² N

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erastova [34]

Answer:

The angular acceleration of the wheel is -6.54 rad/s²

Explanation:

We'll use the equations of motion for this.

w = 2πf

f = 75 rpm = 1.25 rps = 1.25 rev/s

w₀ = initial angular velocity = 2π × 1.25 = 7.85 rad/s

w = final angular velocity = 0 rad/s

t = 1.2 s

α = ?

w = w₀ + αt

0 = 7.85 + 1.2α

α = 7.85/1.2 = - 6.54 rad/s²

6 0
4 years ago
Explain how a transistor behaves as a switch
Zarrin [17]

Answer:

With a zero signal applied to the Base of the transistor it turns “OFF” acting like an open switch and zero collector current flows. With a positive signal applied to the Base of the transistor it turns “ON” acting like a closed switch and maximum circuit current flows through the device.

Explanation:

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5 0
3 years ago
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A concave mirror has a radius of curvature of 1.6m. Find the focal length
Yuki888 [10]
Given,

Radius of curvature of concave mirror = 1.6m

We know that ,

Focal length = radius/2

Hence ,

Focal length of concave mirror = radius of concave mirror /2

=> F = 1.6/2

=> F = 0.8m

Hence the focal length of concave mirror is 0.8 m
5 0
2 years ago
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A rock is thrown upward from a bridge into a river below. The function f(t)=−16t2+44t+88 determines the height of the rock above
babymother [125]

1) 88 ft

2) 4.09 s

3) 1.38 s

4) 118.2 m

Explanation:

1)

For an object thrown upward and subjected to free fall, the height of the object at any time t is given by the suvat equation:

h(t) = h_0 + ut - \frac{1}{2}gt^2 (1)

where

h_0 is the height at time t = 0

u is the initial vertical velocity

g=32 ft/s^2 is the acceleration due to gravity

The function that describes the height of the rock above the surface at a time t in this problem is

f(t)=-16t^2+44t+88 (2)

By comparing the terms with same degree of eq(1) and eq(2), we observe that

h_0 = 88 ft

which means that the rock is at height h = 88 ft when t = 0: therefore, this means that the height of the bridge above the water is 88 feet.

2)

The rock will hit the water when its height becomes zero, so when

f(t)=0

which means when

0=-16t^2+44t+88

First of all, we can simplify the equation by dividing each term by 4:

0=-4t^2+11t+22

This is a second-order equation, so we solve it using the usual formula and we find:

t_{1,2}=\frac{-b \pm \sqrt{b^2-4ac}}{2a}=\frac{-11\pm \sqrt{(11)^2-4(-4)(22)}}{2(-4)}=\frac{-11\pm \sqrt{121+352}}{-8}=\frac{-11\pm 21.75}{-8}

Which gives only one positive solution (we neglect the negative solution since it has no physical meaning):

t = 4.09 s

So, the rock hits the water after 4.09 seconds.

3)

Here we want to find how many seconds after being thrown does the rock reach its maximum height above the water.

For an object in free fall motion, the vertical velocity is given by the expression

v=u-gt

where

u is the initial velocity

g is the acceleration due to gravity

t is the time

The object reaches its maximum height when its velocity changes direction, so when the vertical velocity is zero:

v=0

which means

0=u-gt

Here we have

u=+44 ft/s (initial velocity)

g=32 ft/s^2 (acceleration due to gravity)

Solving for t, we find the time at which this occurs:

t=\frac{u}{g}=\frac{44}{32}=1.38 s

4)

The maximum height of the rock can be calculated by evaluating f(t) at the time the rock reaches the maximum height, so when

t = 1.38 s

The expression that gives the height of the rock at time t is

f(t)=-16t^2+44t+88

Substituting t = 1.38 s, we find:

f(1.38)=-16(1.38)^2 + 44(1.38)+88=118.2 m

So, the maximum height reached by the rock during its motion is

h_{max}=118.2 m

Which means 118.2 m above the water.

3 0
4 years ago
Two hundred turns of (insulated) copper wire are wrapped around a wooden cylindrical core of cross-sectional area 1.20 × 10−3 m2
Novosadov [1.4K]

Answer:

The charge flows through a point in the circuit during the change is 0.044 C.

Explanation:

Given that,

Number of turns in the copper wire, N = 200

Area of cross section, A=1.2\times 10^{-3}\ m^2

Resistance of the circuit, R = 118 ohms

If an externally applied uniform longitudinal magnetic filed in the core changes from 1.65 T in one direction to 1.65 T in the opposite direction.

We need to find the charge flows through a point in the circuit during the change. Due to change in magnetic field an emf is induced in it. It is given by :

\epsilon=\dfrac{d\phi}{dt}

Using Ohm's law :

\epsilon=IR

IR=-\dfrac{d\phi}{dt}\\\\I=-\dfrac{1}{R}\dfrac{d\phi}{dt}

Electric current is equal to the rate of change of electric charge. So,

dq=\dfrac{NA(B(0)-B(t))}{R}\\\\dq=\dfrac{200\times 1.2\times 10^{-3}(1.65+1.65)}{18}\\\\dq=0.044\ C

So, the charge flows through a point in the circuit during the change is 0.044 C.

4 0
3 years ago
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