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uranmaximum [27]
2 years ago
9

A motorcycle travels 3km in five minutes how much distance can travels in one second.

Physics
1 answer:
Zanzabum2 years ago
3 0

10meter

Explanation:

Distance = 3km = 3000Meter

Time = 5 min = 300 sec

distance travel by motorcycle in 5 min = 3km

distance travel by motorcycle in 1sec =

\frac{3000m}{300sec}  = 10meter

.

.

.

Now 30 meter change into km meter

30m = 30/1000 = 0.1km

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An object with a mass of 10 kg is rolled down a frictionless ramp from a height of 3 meters. If a factory worker at the bottom o
ruslelena [56]

Answer:

The amount of work the factory worker must to stop the rolling ramp is 294 joules

Explanation:

The object rolling down the frictionless ramp has the following parameters;

The mass of the object = 10 kg

The height from which the object is rolled = 3 meters

The work done by the factory worker to stop the rolling ramp = The initial potential energy, P.E., of the ramp

Where;

The potential energy P.E. = m × g × h

m = The mass of the ramp = 10 kg

g = The acceleration due to gravity = 9.8 m/s²

h = The height from which the object rolls down = 3 m

Therefore, we have;

P.E. = 10 kg × 9.8 m/s² × 3 m = 294 Joules

The work done by the factory worker to stop the rolling ramp = P.E. = 294 joules

8 0
3 years ago
Nitrogen at 100 kPa and 25oC in a rigid vessel is heated until its pressure is 300 kPa. Calculate (a) the work done and (b) the
nignag [31]

Answer:

A. The work done during the process is W = 0

B. The value of heat transfer during the process Q = 442.83 \frac{KJ}{kg}

Explanation:

Given Data

Initial pressure P_{1} = 100 k pa

Initial temperature T_{1} = 25 degree Celsius = 298 Kelvin

Final pressure P_{2} = 300 k pa

Vessel is rigid so change in volume of the gas is zero. so that initial volume is equal to final volume.

⇒ V_{1} = V_{2} ------------- (1)

Since volume of the gas is constant so pressure of the gas is directly proportional to the temperature of the gas.

⇒ P ∝ T

⇒ \frac{P_{2} }{P_{1}} = \frac{T_{2} }{T_{1}}

⇒ Put all the values in the above formula we get the final temperature

⇒ T_{2} = \frac{300}{100} × 298

⇒ T_{2} = 894 Kelvin

(A). Work done during the process is given by W = P × (V_{2} -V _{1})

From equation (1), V_{1} = V_{2} so work done W = P × 0 = 0

⇒ W = 0

Therefore the work done during the process is zero.

Heat transfer during the process is given by the formula Q = m C_{v} ( T_{2} -T_{1} )

Where m = mass of the gas = 1 kg

C_{v} = specific heat at constant volume of nitrogen = 0.743 \frac{KJ}{kg k}

Thus the heat transfer Q = 1 × 0.743 × ( 894- 298 )

⇒ Q = 442.83 \frac{KJ}{kg}

Therefore the value of heat transfer during the process Q = 442.83 \frac{KJ}{kg}

6 0
3 years ago
"A steel rotating-beam test specimen has an ultimate strength of 120 kpsi. Estimate the life of the specimen if it is tested at
MrRissso [65]

Answer:

life (N) of the specimen is 117000  cycles

Explanation:

given data

ultimate strength Su = 120 kpsi

stress amplitude σa = 70 kpsi

solution

we first calculate the endurance limit of specimen Se i.e

Se = 0.5× Su   .............1

Se = 0.5 × 120

Se = 60 kpsi

and we know strength of friction f  = 0.82

and we take endurance limit Se is = 60 kpsi

so here coefficient value (a) will be

a = \frac{(f\times Su)^2}{Se}     ......................1  

put here value and we get

a = \frac{(0.82\times 120)^2}{60}  

a = 161.4  kpsi

so coefficient value (b) will be

b = -\frac{1}{3}log\frac{(f\times Su)}{Se}  

b =  -\frac{1}{3}log\frac{(0.82\times 120)}{60}  

b = −0.0716

so here number of cycle N will be  

N =  (\frac{ \sigma a}{a})^{1/b}

put here value  and we get

N =  (\frac{ 70}{161.4})^{1/-0.0716}

N = 117000

so life (N) of the specimen is 117000  cycles

7 0
3 years ago
What is the scientific term for rocks formed from lava?
sergeinik [125]

Answer:

extrusive I'm pretty sure that's right

5 0
4 years ago
A wall lizard has adhesive pards in the limbs true or false plz help me​
Leto [7]
True lizards can stick to surfaces because their bulbous toes are covered in hundreds of tiny microscopic hairs called setae
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