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stich3 [128]
3 years ago
9

If an object is moving with constant velocity, what do you know about its acceleration?

Physics
1 answer:
Ivan3 years ago
8 0
It's equals to zero (a=0)
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. A person weighing 750 N gets on an elevator.
Kobotan [32]

 

F = 750 N  (Force)

d = 10 m  (displacement )

t = 25 s   (time)

L = ?   (Mechanical work )  =  (Energy)

P = ?   (Power)

Solve:

L = F × d = 750 × 10 = 7500 Joules

P = L / t = 7500 / 25 = 300 Watts

5 0
2 years ago
The net force acting on an object equals the applied force plus the force of friction.
Georgia [21]

Answer:

False

Explanation:

The net force is equal to the applied force minus the force of friction. It is possible for friction to act in the same direction as an applied force, but that would mean there would have to be more than two forces acting on the object.

3 0
3 years ago
Find the sum of the vectors: 40m/s2 Northeast, 10 m/s2 Northeast
podryga [215]

Answer:

Explanation:

Since both vectors are pointing on the same direction (Northeast), the sum of them will point in that same direction, and its magnitud will be the sum of the magnitudes of each vector (40m/s2+10m/s2). This problem is just a problem in one dimension. The sum of the vectors is then 50m/s2 Northeast.

8 0
3 years ago
An added risk of going to Mars is that there is a lot of radiation in space that can damage tissues. We aren’t sure what will ha
zaharov [31]
I feel like the answer is true
5 0
3 years ago
A lumberjack (mass = 110 kg) is standing at rest on one end of a floating log (mass = 206 kg) that is also at rest. The lumberja
Lemur [1.5K]

Answer:

a). The velocity of the first log is -1.65 m/s.

(b). The velocity of the second log is 1.07 m/s.

Explanation:

Given that,

Mass of lumberjack M= 110 kg

Mass of log m= 206 kg

Final velocity = 3.09 m/s

(a). We need to calculate the velocity of the first log just before the lumberjack jumps off

Using conservation of momentum

Mu_{1}+mu_{2}=Mv_{1}+mv

Put the value into the formula

0=110\times3.09+206v

v=-\dfrac{110\times3.09}{206}

v=-1.65\ m/s

The velocity of the first log is -1.65 m/s.

(b). If the lumberjack comes to rest relative to the second log

We need to calculate the velocity of the second log

(M+m)v=Mv_{1}

v=\dfrac{Mv_{1}}{M+m}

Put the value into the formula

v=\dfrac{110\times3.09}{110+206}

v=1.07\ m/s

The velocity of the second log is 1.07 m/s.

Hence, (a). The velocity of the first log is -1.65 m/s.

(b). The velocity of the second log is 1.07 m/s.

4 0
3 years ago
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