Are surface currents warm or cold? A:war m B:cold

a sprinter with a mass of 80kg accelerates uniformly from 0 m/s to 9 m/s in 3 s. a.)what is the runners acceleration? b.) what i

Katyanochek1 [597]

Part A:
Acceleration can be calculated by dividing the difference of the initial and final velocities by the given time. That is,
a = (Vf - Vi) / t
where a is acceleration,
Vf is final velocity,
Vi is initial velocity, and
t is time

Substituting,
a = (9 m/s - 0 m/s) / 3 s = 3 m/s²
ANSWER: 3 m/s²

Part B:
From Newton's second law of motion, the net force is equal to the product of the mass and acceleration,
F = m x a
where F is force,
m is mass, and
a is acceleration

Substituting,
F = (80 kg) x (3 m/s²) = 240 kg m/s² = 240 N

ANSWER: 240 N 

Part C:
The distance that the sprinter travel is calculated through the equation,
d = V₀t + 0.5at²

Substituting,
d = (0 m/s)(3 s) + 0.5(3 m/s²)(3 s)²
d = 13.5 m

ANSWER: d = 13.5 m

3 0

3 years ago

What process produces radiant energy in stars?

noname [10]

"Nuclear Fusion" produces radiant energy in stars

Hope this helps!

7 0

3 years ago

During her high bar routine from question 2, Gabby Douglas slipped and falls from the high bar, landing on a 10 cm thick gymnast

tankabanditka [31]

a) 122.5 J

b) -122.5 J

c) -1884.6 N

d) -3769.2 N

e) $-753.8 m/s^2$

f) $a=-76.9 g$

Explanation:

a)

The kinetic energy of an object is the energy possessed by the object due to its motion.

Mathematically, it is given by:

$K=\frac{1}{2}mv^2$

where

m is the mass of the object

v is the speed of the object

Here, we want to find the kinetic energy of the head just before hitting the mat.

At that instant, the speed is:

v = 7 m/s

The mass of the head is:

m = 5 kg

So, the kinetic energy is

$K=\frac{1}{2}(5)(7)^2=122.5 J$

b)

According to the work-energy theorem, the work done by a force on an object is equal to the change in kinetic energy of the object:

$W=K_f - K_i$

where

W is the work done

$K_f$ is the final kinetic energy

$K_i$ is the initial kinetic energy

In this problem:

$K_i=122.5 J$ is the kinetic energy of the head just before hitting the mat

$K_f=0 J$ is the final kinetic energy (since the head comes to a stop)

So, the work done by the mat is:

$W=0-122.5 = -122.5 J$

The work is negative because the force exerted by the mat is opposite to the direction of motion of the head.

c)

The work exerted by a force on an object is given by

$W=Fd$

where

F is the force applied

d is the displacement of the object

W is the work done

In this problem:

W = -122.5 J is the work done by the mat on the head

d = 6.5 cm = 0.065 m is the displacement of the head (since it deflects the mat by this amount)

So, the average force exerted by the mat on the head is:

$F=\frac{W}{d}=\frac{-122.5}{0.065}=-1884.6 N$

(the negative sign indicates that the force is in direction opposite to the motion of the head)

d)

The force calculated in part d) represents the average force exerted by the mat on the head:

$F_{avg}=-1884.6 N$

We can assume that as the head first hits the mat, the initial force is zero, then increases at a constant rate up to a peak value of $F_{peak}$ , then it decreases again until the head stops.

In this case, the relationship between average force and peak force is:

$F_{avg}=\frac{0+F_{peak}}{2}$

And therefore, the peak impact force exerted by the mat on the head is:

$F_{peak}=2F_{avg}=2(1884.6)=-3769.2 N$

e)

The peak acceleration of the head can be found by using Newton's second law, which states that:

$F=ma$

where

F is the force on the head

m is the mass of the head

a is the acceleration

Here we have:

F = -3769.2 N is the peak force

m = 5 kg is the mass of the head

So, solving for the acceleration, we find:

$a=\frac{F}{m}=\frac{-3769.2}{5}=-753.8 m/s^2$

f)

The value of the acceleration due to gravity is

$g=9.8 m/s^2$

Here we want to express the peak acceleration of the head in terms of the acceleration due to gravity; so we can write:

$a=Ng$

where

$a=-753.8 m/s^2$ is the peak acceleration

N is the ratio between the peak acceleration and the gravity acceleration

Solving for N,

$N=\frac{a}{g}=\frac{-753.8}{9.8}=-76.9$

This means that the peak acceleration can be written as

$a=-76.9 g$

6 0

3 years ago

Why can work be thought of as the transfer of energy? Please help it's physics

deff fn [24]

That is because work requires energy. According to the law of conservation of energy, it cannot be created or destroyed. When doing work, energy change forms and gets transferred to the object until it is released.
for example, when you lift up an object and place it on a higher elevation, you transferred energy to it and gave it potential energy. The potential energy is transformed into kinetic energy when the object falls down, and if it hits a surface, the energy will scatter, vibrating the areas around it and producing sound.
Also, work= force X distance. The energy does not go away, but rather get changed into some other form of energy

7 0

3 years ago

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How many types air I need a formula of calculating the s i unti of force

yKpoI14uk [10]

The si unit of force is newton.

so, F is eqal to m*g

4 0

3 years ago

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