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Zigmanuir [339]
3 years ago
12

If a sprinter’s mass is 60 kg, how much forward force must be exerted on the sprinter to make the sprinter accelerate at 0.8 m/s

2?
Physics
2 answers:
Natasha2012 [34]3 years ago
7 0
Forward force is 60×0.8
=48N
spin [16.1K]3 years ago
7 0
60 * 0.8 = 48
Thanks (:
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A 45.0-N force pushes a cart 12.5 meters down a hallway. What is the work done on the cart?
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Answer:

562.5J

Explanation:

The following were obtained from the question:

F = 45N

d = 12.5m

w =?

The work done can be achieved by using

w = F x d

w = 45 x 12.5

w = 562.5J

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You are on a boat which is crossing the prime meridian. the altitude of polaris is 50º.explain how you know the boat's location
Lerok [7]
The definition for the prime meridian is: The prime meridian with 0 degrees longitude runs through Greenwich. Polaris is the north star <span>at 50 degrees above the horizon, which means 50 degrees latitude. 
</span>If you observe Polaris  at 50 degrees of altitude, you are at latitude 50 North. You that's why you know that your boat's location is <span>50º north latitude and 0º longitude.</span>
7 0
3 years ago
An astronaut stands by the rim of a crater on the moon, where the acceleration of gravity is 1.62 m/. To determine the depth of
leva [86]

Answer;

=32.15 meters

Explanation;

Use the formula  

s= ut+ 0.5 a (t)^2 to find out 's'  

where s= distance traveled  

u=initial velocity which is zero in this case  

t= time taken to travel 's' distance  

a=acceleration (due to gravity on moon i.e 1.62 m/s^2 )  

Therefore;

S = 0.5 * 1.62 * 6.3 * 6.3

    = 32.1489 meters

Thus; the crater is 32.15 meters deep.

3 0
3 years ago
Which of these scientists had the greatest contribution to early microscopy?
attashe74 [19]

Answer:

b

Explanation:

I think im not really sure tho

7 0
3 years ago
Read 2 more answers
A driver who does not wear a seat belt continues to move at the initial velocity until she or he hits something solid (e.g the s
egoroff_w [7]

This question is incomplete, the complete question is;

Seatbelts provide two main advantages in a car accident (1) they keep you from being thrown from the car and (2) they reduce the force that acts on your during the collision to survivable levels. This second benefit can be illustrated by comparing the net force encountered by a driver in a head-on collision with and without a seat beat.  

1) A driver wearing a seat beat decelerates at roughly the same rate as the car it self. Since many modern cars have a "crumble zone" built into the front of the car, let us assume that the car decelerates of a distance of 1.1 m. What is the net force acting on a 70 kg driver who is driving at 18 m/sec and comes to rest in this distance?

Fwith belt =

2) A driver who does not wear a seat belt continues to move at the initial velocity until she or he hits something solid (e.g the steering wheel) and then comes to rest in a very short distance. Find the net force on a driver without seat belts who comes to rest in 1.1 cm.

Fwithout belt =

Answer:

1) The Net force on the driver with seat belt is 10.3 KN

2) the Net force on the driver without seat belts who comes to rest in 1.1 cm is 1030.9 KN

Explanation:

Given the data in the question;

from the equation of motion, v² = u² + 2as

we solve for a

a = (v² - u²)/2s ----- let this be equation 1

we know that, F = ma ------- let this be equation 2

so from equation 1 and 2

F = m( (v² - u²)/2s )

where m is mass, a is acceleration, u is initial velocity, v is final velocity and s is the displacement.

1)

Wearing sit belt, car decelerates of a distance of 1.1 m. What is the net force acting on a 70 kg driver who is driving at 18 m/sec and comes to rest in this distance.

i.e, m = 70 kg, u = 18 m/s, v = 0 { since it came to rest }, s = 1.1 m

so we substitute the given values into the equation;

F = 70( ((0)² - (18)²) / 2 × 1.1 )

F = 70 × ( -324 / 2.4 )

F = 70 × -147.2727

F = -10309.09 N

F = -10.3 KN

The negative sign indicates that the direction of the force is opposite compared to the direction of the motion.

Fwith belt =  10.3 KN

Therefore, Net force of the driver is 10.3 KN

2)

No sit belt,  

m = 70 kg, u = 18 m/s, v = 0 { since it came to rest }, s = 1.1 cm = 1.1 × 10⁻² m

we substitute

F = 70( ((0)² - (18)²) / 2 × 1.1 × 10⁻² )

F = 70 × ( -324 / 0.022 )

F = 70 × -14727.2727

F = -1030909.08 N

F = -1030.9 KN

The negative sign indicates that the direction of the force is opposite compared to the direction of the motion.

Fwithout belt = 1030.9 KN

Therefore, the net force on the driver without seat belts who comes to rest in 1.1 cm is 1030.9 KN

4 0
3 years ago
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