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PilotLPTM [1.2K]
4 years ago
8

A 40-kg rock is dropped from an elevation of 10 m. What is the velocity of the rock when it is 5-m from the ground?

Physics
1 answer:
ivolga24 [154]4 years ago
6 0

Answer:

Explanation:

Given

mass of rock m=40\ kg

Elevation of Rock h=10\ m

Distance traveled by rock with time

h=ut+\frac{1}{2}at^2

where, u=initial velocity

t=time

a=acceleration

here initial velocity is zero

when rock is 5 m from ground then it has traveled a distance of 5 m from top because total height is 10 m

5=0\times t+\frac{1}{2}(9.8)(t^2)

t^2=\frac{10}{9.8}

t=1.004\approx 1\ s

velocity at this time

v=u+at

v=0+9.8\times 1.004

v=9.83\ m/s

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Liquid nitrogen is kept at a temperature of -320 degrees. When liquid nitrogen is heated it quickly boils and turns into gas. Wh
AysviL [449]

Answer: Picture (d) is the correct answer.

Explanation:

When liquid nitrogen is at -320 degrees then as the temperature is very low, therefore, molecules of nitrogen will be much closer to each other.

Whereas when liquid nitrogen is boiled then we are increasing the temperature. As a result, force of attraction between the molecules decreases and they will move far apart from each other.

It is known that below 63 K liquid nitrogen tends to freeze and thus it becomes solid. So, at -320 degrees liquid nitrogen will be in solid state.

In the given pictures, picture (d) represents the change caused by adding heat to liquid nitrogen.

4 0
3 years ago
Read 2 more answers
A projectile is shot on level ground with a
erastova [34]

Answer:

time rising = 34 / 9.8 = 3.47 sec

total time in air = 2 * 3.47 sec = 6.94 sec

(time rising must equal time falling)

R = 17 m/s * 6.94 s = 118 m

Can also use range formula

R = v^2 sin (2 theta) / g

tan theta = 34 / 17 = 2

theta = 63.4 deg

2 theta = 126.9 deg

sin 126.9 = .8

v^2 = 17^2 + 34^2 = 1445 m^2/s^2

R = 1445 * .8 / 9.8 = 118 m    agreeing with answer found above

5 0
3 years ago
A light spring of constant 163 N/m rests vertically on the bottom of a large beaker of water.
LUCKY_DIMON [66]

Answer:

24.71cm

Explanation:

We approach this problem base don Hooke's law which states the elongation produced in an elastic material is proportional to the applied load or force provided that its elastic limit is not exceeded. This is expressed mathematically as follows;

F=ke................(1)

where F is the applied force, k is the force constant and e is the elongation or extension of the material.

In this problem, the applied force F is the weight of the wood which is calculated as follows,

F=mg.............(2)

m = 4.11kg

g=9.8m/s^2

Hence,

F=4.11*9.8\\F=40.278N

Given that k = 163N/m, we make appropriate substitutions into equation (1) to obtain the following;

40.278=163*e\\e=\frac{40.78}{163}\\e=0.2471m

Since it is required in cm, we perform the conversion as follows, knowing that 100cm = 1m

0.2471m=0.2471*100cm=24.71cm

NB: We do not necessarily need the the density of the wood to perform our calculations since other parameters were given from which we were able to obtain its weight.

7 0
4 years ago
Kinetic energy is measured in joules. True or false
Mademuasel [1]

Answer:True

Explanation:

5 0
3 years ago
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A bicyclist of mass 60kg supplies 340W of power while riding into a 5 m/s head wind. The frontal area of the cyclist and bicycle
NikAS [45]

Answer:

87.1 mph

Explanation:

We are given that

Mass,m=60 kg

Power,P=340 W

Speed,v=5 m/s

Area,A=0.344 m^2

Drag coefficient,C_d=0.88

Coefficient of rolling resistance,\mu_r=0.007

Friction force,f=\mu_rmg=0.007\times 60\times 9.8=4.1 N

Where g=9.8 m/s^2

Let speed of cyclist=v'

Drag force,F_d=\frac{1}{2}\rho_{air}AC_dv^2

Density of air,\rho_{air}=1.225 kg/m^3

F_d=\frac{1}{2}\times 1.225\times 0.344\times 0.88(5)^2=4.635N

Power,P=(F_d+f)\times v'

340=(4.1+4.635) v'=8.735v'

v'=\frac{340}{8.735}=38.9m/s

v'=87.1 mph

1 m=0.00062137 miles

1 hour=3600 s

7 0
4 years ago
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