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victus00 [196]
3 years ago
9

What information is represented by the slope of a graph of displacement versus time (displacement-time graph)?​

Physics
1 answer:
OverLord2011 [107]3 years ago
7 0
C - velocity

When you look at a point it is saying it goes this many meters in this amount of time. This is speed which means the slope is velocity
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A boy carries a 50 N box up a flight of stairs 10-m high. How much potential energy has the box gained?
almond37 [142]

Answer:

Potential energy, P = 500 J

Explanation:

We have,

Weight of the box, W = 50 N

Height of the stairs, h = 10 m

It is required to find the potential energy gained by the box. The potential energy is possessed by an object due to its height. The formula of potential energy is given by :

P=mgh

g is acceleration due to gravity

P=50\ N\times 10\ m\\\\P=500\ J

So, the potential energy of 500 J is gained by the box.

3 0
3 years ago
If net force is zero, the what will be constant.
vodomira [7]

Answer:

The ans is velocity

Explanation:

Hope it helps you

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3 years ago
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Which of the following particles is NOT a candidate for dark matter?
murzikaleks [220]

its C) Axion                        

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Give one real-life example of the Doppler Effect and explain what causes this phenomenon. Please write at least 3 complete sente
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3 years ago
A 250-kN railroad car A is traveling at 40 m/s while a 550-kN freight car B is traveling at 10 m/s (both cars heading to the rig
Damm [24]

Answer:

the maximum deformation undergone by the spring = 47.46 cm

Explanation:

Using conservation of  momentum:

m_Av_A + m_Bv_B = (m_A+m_B )v

where:

m_A = \frac{250\ kN}{g}

v_A = 40

m_B = \frac{550\ kN}{g}

v_B =10

Then;

m_Av_A + m_Bv_B = (m_A+m_B )v

\frac{250*10^3}{9.81}*40 + \frac{550*10^3}{9.81}*10 = (\frac{800*10^3}{9.81} )v

1580020.387 = 81549.43935 \ v

v = \frac{1580020.387}{81549.43935}

v = 19.375 m/s

However ; using conservation of energy to determine the maximum deformation undergone by the spring ; we have:

\frac{1}{2} [m_Av_A^2 +m_Bv_B^2] =\frac{1}{2}[(m_A+m_B)v^2 + kx^2]

[m_Av_A^2 +m_Bv_B^2] =[(m_A+m_B)v^2 + kx^2]

[\frac{250*10^3}{9.81}*40^2 + \frac{550*10^3}{9.81}*10^2] =[ (\frac{800*10^3}{9.81} )*19.375^2 + 70 *10^6 \ * x^2]

x = 0.4746 m

x = 47.46 cm

Thus,  the maximum deformation undergone by the spring = 47.46 cm

3 0
4 years ago
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