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Vera_Pavlovna [14]
1 year ago
14

A student who was training for a cross country race jogged for 2. 0 hours and covered a distance of 14. 0 kilometers. What was t

he average speed of the student?.
Physics
1 answer:
FinnZ [79.3K]1 year ago
6 0

The average speed is 116.66m/s.

Given - The path traced is 14km , time for jogging is 2 hrs=120min

To find the average speed-

  1. Speed refers to the ease of the movement and degree of mobility as a result of force application.
  2. Due to this there is involvement of velocity.
  3. Journey of average speed is the cumulative of distances and time.
  4. Kinetic theory refers to the Boltzmann constant connecting to the standards of distance traversed.

calculations-

Speed= \frac{Distance}{Time}

Speed= 14 000 / 120

= 116.66m/s

To learn more about average speed -

<u>brainly.com/question/27753148</u>

#SPJ4

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Ice-skater slides toward a sled sitting on the ice and hits it. The skater exerts a 12.6 N force on the sled at an angle of 15.3
RideAnS [48]

Answer:

Expression of work done is

W = Fd cos\theta

Work done to move the sled is given as 187.2 J

Explanation:

As we know that the formula of work done is given as

W = Fd cos\theta

here we know that

F = 12.6 N

d = 15.4 m

\theta = 15.3 degree

so we will have

W = 12.6 \times 15.4 cos15.3

W = 187.2 J

5 0
3 years ago
5g of ammonium nitrate was dissolved in 60g of water in an insulated container. The temperature at the start of the reaction was
Minchanka [31]

Answer: The energy absorbed by the reaction from the water is 996 Joules.

Explanation:

Energy absorbed by the reaction or energy lost by the water to the reaction,Q.

Mass of the the reaction  ,m = 60 g

Specific heat of water = c = 4.15 J\g ^oC

Change is temperature=\Delta T=19^oC-23^oC=-4^oC

Q=mc\Delta T=60 g\times 4.15 J\g ^oC\times (-4^oC)=-996 Joules

Negative sigh indicates that energy was given by the water to the reaction.

The energy absorbed by the reaction from the water is 996 Joules.

5 0
3 years ago
Ten identical steel wires have equal lengths L and equal "spring constants" k. The Young's modulus of each wire is Y. The wires
svlad2 [7]

Answer:

option (B)

Explanation:

Young's modulus is defined as the ratio of longitudinal stress to the longitudinal strain.

Its unit is N/m².

The formula for the Young's modulus is given by

Y=\frac{F \times L}{A\times \delta L}

where, F is the force applied on a rod, L is the initial length of the rod, ΔL is the change in length of the rod as the force is applied, A is the area of crossection of the rod.

It is the property of material of solid. So, when the 10 wires are co joined together to form a new wire of length 10 L, the material remains same so the young' modulus remains same.

8 0
3 years ago
One end of an insulated metal rod is maintained at 100 ∘C and the other end is maintained at 0.00 ∘C by an ice–water mixture. Th
lozanna [386]

Answer:

k=105.0359\times 10^4\,W.m^{-1}.K^{-1}

Explanation:

Given:

temperature at the hotter end, T_H=100^{\circ}C

temperature at the cooler end, T_C=0^{\circ}C

length of rod through which the heat travels, dx=0.7\,m

cross-sectional area of rod, A=1.1\times 10^{-4}\,cm^2

mass of ice melted at zero degree Celsius, m=8.7\times 10^{-3}\,kg

time taken for the melting of ice, t=15\times60=900\,s

thermal conductivity k=?

By Fourier's Law of conduction we have:

\dot{Q}=k.A.\frac{dT}{dx}......................................(1)

where:

\dot{Q}=rate of heat transfer

dT= temperature difference across the length dx

Now, we need the total heat transfer according to the condition:

we know the latent heat of fusion of ice,  L = 334000\,J.kg^{-1}

Q=m.L

Q=8.7\times 10^{-3}\times 334000

Q=2905.8\,J

Now the heat rate:

\dot{Q}=\frac{Q}{t}

\dot{Q}=\frac{2905.8}{900}

\dot{Q}=3.2287\,W

Now using eq,(1)

3.2287=k\times 1.1\times 10^{-4} \times \frac{100}{0.7}

k=205.4606\,W.m^{-1}.K^{-1}

8 0
3 years ago
Block b rests upon a smooth surface. if the coefficients of static and kinetic friction between a and b are μs = 0.4 and μk = 0.
aliina [53]

Given

Weight of the block A, Wa = 20 lb, weight of block B Wb = 50 lb. Applied force to block A, P = 6lb, coefficient of static friction µs = 0.4, coefficient of kinetic friction µk = 0.3. If a force P is applied to the body, no relative motion will take place until the applied force is equal to the force of friction Ff, which is acting opposite to the direction of motion. Magnitude of static force of friction between block A and block B, Fs = µsN, where N is reaction force acting on block A. Now, resolve the forces Fx = max. P = (mA + mB)a,

 

6 = (20 / 32.2 + 50 / 32.2)a

 

2.173a = 6

 

A = 2.76 ft/s^2

 

To check slipping occurs between block A and block B, consider block A:

P – Ff = mAaA

6 – Ff = 1.71

Ff = 4.29 lb

 

And also,

N = wA. We know static friction,

Fs = µsN

Fs = 0.4 x 20

Fs = 8lb

Frictional force is less than static friction. Ff < Fs

<span>Therefors, acceleration of block A, aA = 2.76 ft/s^2, acceleration of block B aB = 2.76 ft/s^2</span>

6 0
3 years ago
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