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djverab [1.8K]
3 years ago
12

a man stands in a lift going downward with uniform velocity. he experiences a loss of weight at the start but not when lift is i

n uniform motion ..explain why​
Physics
1 answer:
AnnyKZ [126]3 years ago
6 0

Answer:

It is explained in the explanation section

Explanation:

When the lift starts going downwards, it will start accelerating downwards. After a while, it will start moving with a constant velocity.

Constant velocity means that acceleration is zero and so the man will not feel any weight loss.

Now, Once the lift achieves constant velocity the acceleration is zero hence he will not experience any weight loss.

However, when the lift is in uniform motion, the lift and the man will fall down with an acceleration(a) that is less than that due to gravity(g) . Thus, the man will feel an apparent weight F which is not equal to zero.

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What does the area between the line and the x-axis represent on a velocity vs. time graph?
Finger [1]

Answer:

It has been learned in this lesson that the area bounded by the line and the axes of a velocity-time graph is equal to the displacement of an object during that particular time period. ... Once calculated, this area represents the displacement of the object.

Explanation:

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3 years ago
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Today’s scientist must search through scientific journals before performing an experiment with making methodical observations wh
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Answer:

in the lab

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cause that is where scientist spend their time doing research ...

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3 years ago
A ball rolls horizontally off a table and a height of 1.4 m with a speed of 4 m/s. How long does it take the ball to reach the g
Hitman42 [59]

For vertical motion, use the following kinematics equation:

H(t) = X + Vt + 0.5At²

H(t) is the height of the ball at any point in time t for t ≥ 0s

X is the initial height

V is the initial vertical velocity

A is the constant vertical acceleration

Given values:

X = 1.4m

V = 0m/s (starting from free fall)

A = -9.81m/s² (downward acceleration due to gravity near the earth's surface)

Plug in these values to get H(t):

H(t) = 1.4 + 0t - 4.905t²

H(t) = 1.4 - 4.905t²

We want to calculate when the ball hits the ground, i.e. find a time t when H(t) = 0m, so let us substitute H(t) = 0 into the equation and solve for t:

1.4 - 4.905t² = 0

4.905t² = 1.4

t² = 0.2854

t = ±0.5342s

Reject t = -0.5342s because this doesn't make sense within the context of the problem (we only let t ≥ 0s for the ball's motion H(t))

t = 0.53s

8 0
3 years ago
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Which statement is true for a car that first goes around a curve of radius r at a constant speed v, and then goes around the sam
sasho [114]
Centripetal force is equal to (mv^2)/r
The way I use to answer these question is to set every variable to 1
m=1
v=1
r=1
so centripetal force =1
then change the variable we're looking at
and since we're find when it's half we could either change it to 1/2 or 2, but 2 is easier to use
m=1
v=2
r=1
((1)×(2)^2)/1=4
So the velocity in the 1st part is half the velocity in the 2nd part and the centripetal force is 4× less
The answer is the centripetal force is 1/4 as big the second time around
3 0
3 years ago
A neutron star and a black hole are 3.34 x 1012 m from each other at a certain point in their orbit. The neutron star has a mass
m_a_m_a [10]

Answer:

  F=1.65 x 10²⁶ N

Explanation:

Given that

Distance ,R= 3.34 x 10¹² m

Mass m₁= 2.78 x 10³⁰ kg

Mass ,m₂= 9.94 x 10³⁰ kg

we know that gravitational force F given as

F=G\dfrac{m_1m_2}{R^2}

G=Constant

G=6.67 x 10⁻¹¹ Nm²/kg²

Now by putting the values

F=6.67\times 10^{-11}\times \dfrac{2.78\times 10^{30}\times 9.94\times 10^{30}}{(3.34\times 10^{12})^2}\ N

F=1.65 x 10²⁶ N

Therefore the force between these two mass will be 1.65 x 10²⁶ N.

5 0
3 years ago
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