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dmitriy555 [2]
2 years ago
15

How much work must be done to raise a 1100kg car 2m above the ground?​

Physics
2 answers:
N76 [4]2 years ago
6 0
21560J? i dont know
Xelga [282]2 years ago
4 0

Answer:

21560 J

Explanation:

Work = mg*h = 1100*9.8*2 = 21560 J

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The best answer is D. field lines should always be crossing each other.
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A bat strikes a 0.145-kg baseball. Just before impact, the ball is traveling horizontally to the right at 60.0 m/s , and it leav
nata0808 [166]

Answer:

F = -307.4 N

Explanation:

It is given that,

Mass of the baseball, m = 0.145 kg

Initial speed of the baseball, u = 60 m/s

Final speed of the baseball, v=65\ cos(30)=56.29\ m/s

Time of contact, t=1.75\ ms=1.75\times 10^{-3}\ s

(a) It is assumed to find the horizontal component of average force. It is given by :

F=m\dfrac{v-u}{t}

F=0.145\dfrac{56.29-60}{1.75\times 10^{-3}}

F = -307.4 N

So, the horizontal component of average force is 307.4 N. Hence, this is the required solution.

8 0
3 years ago
"a wind shift from the south or southwest to the northwest is commonly associated with the passage of which type of front
cluponka [151]
If i wouldve know i would tell you sorry
4 0
2 years ago
O'Malley is riding on a bus which is moving at 10 m/s, and he throws a ball which he observes to be moving at 10 m/s relative to
Vikki [24]

Answer:

<em>20 m/s in the same direction of the bus.</em>

Explanation:

<u>Relative Motion </u>

Objects movement is always related to some reference. If you are moving at a constant speed, all the objects moving with you seem to be at rest from your reference, but they are moving at the same speed as you by an external observer.

If we are riding on a bus at 10 m/s and throw a ball which we see moving at 10 m/s in our same direction, then an external observer (called Ophelia) will see the ball moving at our speed plus the relative speed with respect to us, that is, at 20 m/s in the same direction of the bus.

3 0
2 years ago
A police officer is parked by the side of the road, when a speeding car travelling at 50 mi/hrpasses. The police car immediately
Blababa [14]

Answer:

a) time taken to catch up with speeding car is 12.25 secs

b) the police car will travel 273.8 m to catch up with the speeding car

Explanation:

Given that;

speed of car V_{c} = 50 mi/hr = 22.352 m/s

acceleration of police car = 10 mi/hr = 4.47 m/s²

V_{f}  = 70 mi/hr = 31.29 m/s

Now time taken to reach maximum speed is t₁

so

V_{f} =  V_{i} + at₁

we substitute

31.29 = 0 + 4.47t₁

t₁ = 31.29 / 4.47

t₁  = 7 sec

now

d₁ = 0 + 1/2 × at₁²

d₁ = 0 + 1/2 × 0 + 4.47×(7)²

d₁ = 109.5 m

so distance travelled by the speeding car in time t₁  will be

d_{c} = V_{c} × t₁

we substitute

d_{c} = 22.352 × 7

d_{c}  = 156.46 m

now distance between polive car and speeding car

Δd =  d_{c} - d₁

Δd = 156.46 - 109.5

Δd = 46.96 m

time taken to cover Δd will be

t₂ = Δd / ( V_{f} - V_{c} )

t₂ = 46.96 / ( 31.29 - 22.352 )

t₂ = 46.96 / 8.938

t₂ = 5.25 sec

distance travelled by the police in time t₂ will be

d₂ = V_{f} × t₂

d₂ = 31.29 × 5.25

d₂ = 164.3 m

a) How long will it take before the officer catches up to the speeding car;

time taken to catch up with speeding car;

t = t₁ + t₂

t = 7 + 5.25

t = 12.25 secs

Therefore, time taken to catch up with speeding car is 12.25 secs

b)  how far will it have travelled in order to do so;

distance = d₁ + d₂

distance = 109.5 + 164.3

distance = 273.8 m

Therefore, the police car will travel 273.8 m to catch up with the speeding car

6 0
3 years ago
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