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UkoKoshka [18]
3 years ago
10

What is the period of an objects motion?

Physics
1 answer:
Lyrx [107]3 years ago
8 0
The time for an object to complete one full cycle. Can have a long period or short period.


Brainliest?
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Consider the collision of a 1500-kg car traveling east at 20.0 m/s (44.7 mph) with a 2000-kg truck traveling north at 25 m/s (55
masha68 [24]

Answer:

1. X_{cm}=-8.57m

2. Y_{cm}=-14.29m

3. V_{cm} = 16.66m/s

4. α = 59.05°

5. V_{cm} = 16.66m/s

6. α = 59.05°

Explanation:

The position of the center of mass 1s before the collision is:

X_{cm}=\frac{X_c*mc+X_t*m_t}{m_c+m_t}

where

X_c=-20m;  m_c=1500kg;  

X_t=0m;   m_t=2000kg;

Replacing these values:

X_{cm}=-8.57m

Y_{cm}=\frac{Y_c*mc+Y_t*m_t}{m_c+m_t}

where

Y_c=0m;  m_c=1500kg;  

Y_t=-25m;   m_t=2000kg;

Replacing these values:

Y_{cm}=-14.29m

The velocity of their center of mass is:

V_{cm-x}=\frac{V_{c-x}*mc+V_{t-x}*m_t}{m_c+m_t}

where

V_{c-x}=20m/s;  m_c=1500kg;  

V_{t-x}=0m/s;   m_t=2000kg;

Replacing these values:

V_{cm-x}=8.57m/s

V_{cm-y}=\frac{V_{c-y}*mc+V_{t-y}*m_t}{m_c+m_t}

where

V_{c-y}=0m;  m_c=1500kg;  

V_{t-y}=-25m;   m_t=2000kg;

Replacing these values:

V_{cm-y}=-14.29m

So, the magnitude of the velocity is:

V_{cm}=\sqrt{V_{cm-x}^2+V_{cm-y}^2}

V_{cm}=16.66m/s

The angle of the velocity is:

\alpha =atan(V_{cm-y}/V_{cm-x})

\alpha=59.05\°

Since on any collision, the velocity of the center of mass is preserved, then the velocity after the collision is the same as the previously calculated value of 16.66m/s at 59.0° due north of east

4 0
3 years ago
Let’s allow ourselves to be a little more optimistic with our bus passengers. Earlier in the movie Sandy has to make the bus jum
Roman55 [17]

Answer:

25.71486 seconds

Explanation:

t = Time taken

u = Initial velocity = 60 mph

v = Final velocity = 80 mph

s = Displacement = 0.5 mile

a = Acceleration

Converting to m/s

60\times \dfrac{1609.34}{3600}=26.8223\ m/s

80\times \dfrac{1609.34}{3600}=35.7631\ m/s

From equation of motion

v^2-u^2=2as\\\Rightarrow a=\dfrac{v^2-u^2}{2s}\\\Rightarrow a=\dfrac{35.7631^2-26.8223^2}{2\times \dfrac{1609.34}{2}}\\\Rightarrow a=0.34769\ m/s^2

v=u+at\\\Rightarrow t=\dfrac{v-u}{a}\\\Rightarrow t=\dfrac{35.7631-26.8223}{0.34769}\\\Rightarrow t=25.71486\ s

Time taken is 25.71486 seconds

4 0
4 years ago
The temperature of a sample of silver increased by 24.0 °C when 269 J of heat was applied. What is the mass of the sample?
statuscvo [17]

Answer:

Mass of the silver will be equal to 46.70 gram

Explanation:

We have given heat required to raise the temperature of silver by 24°C is 269 J , so \Delta T=24^{\circ}C

Specific heat of silver = 0.240 J/gram°C

We have to find the mass of silver

We know that heat required is given by

Q=mc\Delta T, here m is mass, c is specific heat of silver and \Delta T is rise in temperature

So 269=m\times 0.240\times 24

m = 46.70 gram

So mass of the silver will be equal to 46.70 gram

3 0
3 years ago
True or False?<br><br> There are equal numbers of protons and electrons in neutral atoms.
Alecsey [184]
Answer: True
Hope this helped
8 0
3 years ago
Read 2 more answers
Calculate the acceleration of a mobile that at 4s is 32m from the origin, knowing that its initial speed is 10m / s.
ehidna [41]

Answer:

5.5 m/s^2

Explanation:

I believe this is the answer > using the formula a= v-v0/t

Hope this helps!

7 0
3 years ago
Read 2 more answers
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