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Paraphin [41]
3 years ago
10

A runner begins running at the beginning of 7th period and stops running at the end of the period. She runs at a pace of 10 km/h

r and travels 9.2 km. How long did she run
Physics
1 answer:
MrRissso [65]3 years ago
8 0

Answer:

She run for, t = 0.92 s

Explanation:

Given data,

The velocity of the runner, v = 10 km/h

The distance covered by the runner, d = 9.2 km

The relationship between the velocity, displacement and time is given by the formula,

                           t = d / v

Substituting the given values in the above equation,

                           t = 9.2 / 10

                             = 0.92 s

Hence, she ran for, t = 0.92 s

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N april 1974, steve prefontaine completed a 10-km race in a time of 27 min, 43.6 s. suppose \"pre\" was at the 7.15-km mark at a
aliya0001 [1]

solution:speed is 7.99km ,u=\frac{7990}{(25\times60)}=5.33m/s\\
now let the final velocity after 60s from 7.99km is v.\\
v-u =at\\v=5.33+a\times60\\  =60s+5.33\\now distance travelled during this time will be\\
s_{2}=vt\\
v\times103.6\\
=(60a+5.33)\times103.6\\
=551.84+6216am\\
now total distance travelled  is \\  also\\
s_{1}=ut+(\frac{1}{2})at^{2}      =5.33\times60+(0.5)9\times3600\\
=1800a+319.6m\\
Now the remaining time to complete rest race was\\
t=(27\times60+43.6)60-60(25\times60)\\
=103.6s\\Now, distance travelled during this time will be\\
s_{2}=vt\\
v\times103.6\\
=(60a+5.33)\times103.6\\
=551.84+6216am\\
now total distance travelled  is \\
s_{1}+s_{2}+7990=10000\\
1800a+319.6+551.84+6216a=2010\\
8016a=2010-871.44\\
8016a=1138.56\\
a=0.142m/s^2

6 0
3 years ago
Help please, It's for science :>
xenn [34]

Answer:

he tail of the arrow moves a distance of 0.5 m as the arrow is shot. yare yare daze

Explanation:

3 0
3 years ago
Read 2 more answers
How long will it take a 2.3"x10^3 kg truck to go from 22.2 m/s to a complete stop if acted on by a force of -1.26x10^4 N.What wo
Greeley [361]

The stopping distance is 45.0 m

Explanation:

First of all, we find the acceleration of the truck, by using Newton's second law:

F=ma

where

F=-1.26\cdot 10^4 N is the net force on the truck

m=2.3\cdot 10^3 kg is the mass of the truck

a is its acceleration

Solving for a,

a=\frac{F}{m}=\frac{-1.26\cdot 10^4}{2.3\cdot 10^3}=-5.48 m/s^2

where the negative sign means the acceleration is opposite to the direction of motion.

Now, since the motion of the truck is at constant acceleration, we can apply the following suvat equation:

v^2-u^2=2as

where

v = 0 is the final velocity of the truck

u = 22.2 m/s is the initial velocity

a=-5.48 m/s^2 is the acceleration

s is the stopping distance

And solving for s,

s=\frac{v^2-u^2}{2a}=\frac{0-(22.2)^2}{2(-5.48)}=45.0 m

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6 0
4 years ago
One ounce of a well-known breakfast cereal contains 103 Calories (1 food Calorie = 4186 J). If 1.6% of this energy could be conv
faust18 [17]

Answer:

the heaviest barbell that could be lifted  is 390.6kg

Explanation:

Hello!

To solve this question you must follow the following steps

1. Find the amount of energy that can transform the body into motion, this is achieved by multiplying the breakfast consumed by the percentage of energy conversion.

W=(103Calories)(0.016)\frac{4186J}{1calorie} =6898.5J

2. We use the equation that defines the work done by a body that has weight when it is lifted, this is defined by the product of mass by gravity by height.

W=mgh

where

W=work=6898.5J

m=mass

g=gravity=9.81m/s^2

h=height=1.8m

now we solve for mass, and use the values.

m=\frac{W}{hg} =\frac{6898.5}{(1.8)(9.81)} =390.6kg

the heaviest barbell that could be lifted  is 390.6kg

5 0
3 years ago
A kangaroo has a maximum gravitational potential energy during one jump of 770 J
KengaRu [80]

Answer:

k = 44000 N/m

Explanation:

Given the following data;

Maximum gravitational potential energy = 770 J

Elastic potential energy = 14% of 770 J = 14/100 * 770 = 107.8 J

Extension, x = 42 - 35 = 7cm to meters = 7/100 = 0.07 m

To find the spring constant, k;

The elastic potential energy of an object is given by the formula;

E.P.E = \frac {1}{2}kx^{2}

Substituting into the equation, we have;

107.8 = \frac {1}{2}*k*0.07^{2}

107.8 = \frac {1}{2}*k*0.0049

Cross-multiplying, we have;

215.6 = k*0.0049

k = \frac {215.6}{0.0049}

k = 44000 N/m

5 0
3 years ago
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