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poizon [28]
3 years ago
15

A 2kg object initially going 4m/s to the right is later going 8m/s. whats the change in velocity?

Physics
1 answer:
Nezavi [6.7K]3 years ago
6 0

The change in velocity is +4 m/s to the right (or -4 m/s to the left).

The object's mass is irrelevant.

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A 20kg mass approaches a spring at a speed of 30 m/s. The mass compresses the spring 12cm before coming to a stop. Calculate the
Oksana_A [137]

Answer:

625000 N/ m

Explanation:

m= 20 kg

v= 30 m/s

x= 12 cm

k = ?

Here when the mass when hits at spring its speed is

Vi= 30 m/s

Finally it comes to rest after compressing for 12 cm

i-e Vf = 0 m/s

Distance= S= 12 cm = 0.12 m

using

2aS= Vf2 - Vi2

==> 2a ×0.12 = o- 30 × 30

==> a = 900 ÷ 0.24 = 3750 m/sec2

Now we know;

F = ma

F= -Kx

==> ma= -kx

==> 20 × 3750 = -K × 0.12

==> k = 625000 N/ m

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3 years ago
A 103 kg horizontal platform is a uniform disk of radius 1.71 m and can rotate about the vertical axis through its center. A 68.
Andreyy89

Answer:

I_{total}=220.64 kg*m^{2}

Explanation:

The moment of inertia of the system is equal to the each population and the platform inertia so

Inertia disk

I_{disk}=\frac{1}{2}*m_{disk}*(r_{p})^{2}

Inertia person

I_{p}=\frac{1}{2}*m_{p}*(r_{p})^{2}

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The Inertia of the system is the sum of each mass taking into account that all exert the force of inertia:

I_{total}=I_{disk}+I_{p}+I_{d}

I_{total}=\frac{1}{2}*103kg*(1.71)^{2}+\frac{1}{2}*68.9kg*(1.09)^{2}+\frac{1}{2}*27.7kg*(1.45)^{2}

I_{total}=220.64 kg*m^{2}

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3 years ago
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Hello!

\large\boxed{p = 5  \text{ } kgm/s}

Use the equation for momentum:

p = mv

Plug in the given mass and velocity into the equation:

p = 10 * 0.5

p = 5 kgm/ s

4 0
3 years ago
Which point on the graph shows a temperature of 40°C and the time of 25 minutes?
const2013 [10]

Answer:

'A' is the the point on the graph that shows a temperature of 40°C and the time of 25 minutes

8 0
4 years ago
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