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3 years ago

What is the sprinters output at 2.0 s, 4.0 s and 6.0 s?

Physics

1 answer:

kari74 [83]3 years ago

5 0

A 59 kg sprinter, starting from rest, runs 47 m in 7.0 s at constant acceleration.?

What is the sprinter's power output at 2.0 s, 4.0 s, and 6.0 s?

Instantaneous Power is the force times velocity

P = Fv

Because the acceleration is constant, the force will be constant as well

F = ma

P = mav

for constant acceleration, the velocity at each time is found using

v = at

P = ma(at) = ma²t

find the acceleration using kinematic equation

s = ½at²

a = 2s/t²

a = 2(47) / 7.0²

a = 1.918 m/s²

P(2.0) = 59(1.918²)2.0 = 434.25 W = 0.43 kW

P(4.0) = 59(1.918²)4.0 = 868.51 W = 0.87 kW

P(6.0) = 59(1.918²)6.0 = 1302.76 W = 1.3 kW

I hope this helped.

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The relationship among speed, distance, and time is

alexdok [17]

Distance= speed (multiplied by) time

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2 years ago

Read 2 more answers

A 0.274 kg block is pushed up a frictionless ramp inclined at angle of 32° above the horizon. The push force is a constant 2.55

SpyIntel [72]

Answer:

The answer is 2,416 m/s. Let's jump in.

Explanation:

We do work with the amount of energy we can transfer to objects. According to energy theory:

W = ΔE

Also as we know W = F.x

We choose our reference point as a horizontal line at the block's rest point.<u> At the rest, block doesn't have kinetic energy</u> and <u>since it is on the reference point(as we decided) it also has no potential energy.</u>

Under the force block gains;

W = F.x → $W=2,55.0,71=1,8105\frac{N}{m}$

In the second position block has both kinetic and potential energy. Following the law of conservation of energy;

W = ΔE = Kinetic energy + Potantial Energy

W = ΔE = $\frac{1}{2} mV^{2} + mgh$

Here we can find h in the triangle i draw in the picture using sine theorem;

In a triangle $\frac{a}{sinA}=\frac{b}{sinB}=\frac{c}{sinC}$

In our situation

$\frac{0,71}{sin90} =\frac{h}{sin32}$ → $h=0,376$

Therefore

$1,8105=\frac{1}{2} 0,274V^{2} +0,274.9,81.0,376$

→ $V=2,416$

7 0

3 years ago

Three masses (3 kg, 5 kg, and 7 kg) are located in the xy-plane at the origin, (2.3 m, 0), and (0, 1.5 m), respectively.

Artist 52 [7]

Answer:

a) C.M $=(\bar x, \bar y)=(0.767,0.7)m$

b) $(x_4,y_4)=(-1.917,-1.75)m$

Explanation:

The center of mass "represent the unique point in an object or system which can be used to describe the system's response to external forces and torques"

The center of mass on a two dimensional plane is defined with the following formulas:

$\bar x =\frac{\sum_{i=1}^N m_i x_i}{M}$

$\bar y =\frac{\sum_{i=1}^N m_i y_i}{M}$

Where M represent the sum of all the masses on the system.

And the center of mass C.M $=(\bar x, \bar y)$

Part a

$m_1= 3 kg, m_2=5kg,m_3=7kg$ represent the masses.

$(x_1,y_1)=(0,0),(x_2,y_2)=(2.3,0),(x_3,y_3)=(0,1.5)$ represent the coordinates for the masses with the units on meters.

So we have everything in order to find the center of mass, if we begin with the x coordinate we have:

$\bar x =\frac{(3kg*0m)+(5kg*2.3m)+(7kg*0m)}{3kg+5kg+7kg}=0.767m$

$\bar y =\frac{(3kg*0m)+(5kg*0m)+(7kg*1.5m)}{3kg+5kg+7kg}=0.7m$

C.M $=(\bar x, \bar y)=(0.767,0.7)m$

Part b

For this case we have an additional mass $m_4=6kg$ and we know that the resulting new center of mass it at the origin C.M $=(\bar x, \bar y)=(0,0)m$ and we want to find the location for this new particle. Let the coordinates for this new particle given by (a,b)

$\bar x =\frac{(3kg*0m)+(5kg*2.3m)+(7kg*0m)+(6kg*a)}{3kg+5kg+7kg+6kg}=0m$

If we solve for a we got:

$(3kg*0m)+(5kg*2.3m)+(7kg*0m)+(6kg*a)=0$

$a=-\frac{(5kg*2.3m)}{6kg}=-1.917m$

$\bar y =\frac{(3kg*0m)+(5kg*0m)+(7kg*1.5m)+(6kg*b)}{3kg+5kg+7kg+6kg}=0m$

$(3kg*0m)+(5kg*0m)+(7kg*1.5m)+(6kg*b)=0$

And solving for b we got:

$b=-\frac{(7kg*1.5m)}{6kg}=-1.75m$

So the coordinates for this new particle are:

$(x_4,y_4)=(-1.917,-1.75)m$

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2 years ago

The universal force that is effective over the longest distance is called?

gulaghasi [49]

The answer is D. gravity

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3 years ago

How many possible outcomes are there in rolling a spinner 1 to 8?; What is the probability of spin an even number on a 1 to 8 sp

UkoKoshka [18]

The probability of getting a prime number ( 2, 3, 5, 7) between 1 to 8 is 1/2. The total outcomes possible for spinning the spinner and tossing the coin is 16.

<h3>What is probability?</h3>

Branch of mathematics concerning numerical descriptions of how likely an event is to occur is called probability.

Total number of outcomes = 8.

Let, the probability of getting a prime number ( 2, 3, 5, 7) between 1 to 8

P(E) = (no. of outcomes) /( total no. of outcomes)

=4/8

=1/2

When a coin is tossed, there are two possible outcomes that are H(head), and T(tail).

If you spin the spinner , there are 8 possibilities.

Total outcomes = 8 * 2

=16

There are 16 outcomes, when spinning a spinner numbered from 1 to 8 and tossing a coin.

To know more about probability, refer

brainly.com/question/13604758

#SPJ4

7 0

1 year ago