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guajiro [1.7K]
3 years ago
11

Mary travelled 70 miles/hour due north.

Physics
2 answers:
slega [8]3 years ago
7 0

Answer : The correct option is, Velocity.

Explanation :

Acceleration : It is defined as the rate of change of velocity of an object wit respect to the time. Acceleration is a vector quantity that means it depends on the magnitude and direction of an object. The S.I unit of acceleration is, m/s^2

Distance : It is defined as the how far an object has traveled in time. Distance is a scalar quantity that means it is depends on the magnitude of an object only. The S.I unit of distance is, m

Speed : It is defined as the distance traveled by an object in unit time. Speed is also a scalar quantity. The S.I unit of speed is, m/s

Velocity : It is defined as the change in the position with respect to the time. Velocity is a vector quantity that means it depends on the magnitude and direction of an object. The S.I unit of velocity is, m/s

Mary traveled 70 miles/hour due north. This is an example of velocity. 70 miles/hour tell us about the magnitude of the object and north tell us about the direction of an object.

Hence, the correct option is, velocity.

kap26 [50]3 years ago
3 0
Since we know Mary's speed AND direction, we have
all the information we need to state her velocity.
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An experiment is done to compare the initial speed of projectiles fired from high-performance catapults. The catapults are place
anzhelika [568]

Answer:1.084

Explanation:

Given

mass of Pendulum M=10 kg

mass of bullet m=5.5 gm

velocity of bullet u

After collision let say velocity is v

conserving momentum we get

mu=(M+m)v

v=\frac{m}{M+m}\times u

Conserving Energy for Pendulum

Kinetic Energy=Potential Energy

\frac{(M+m)v^2}{2}=(M+m)gh

here h=L(1-\cos \theta ) from diagram

therefore

v=\sqrt{2gL(1-\cos \theta )}

initial velocity in terms of v

u=\frac{M+m}{m}\times \sqrt{2gL(1-\cos \theta )}

For first case \theta =6.8^{\circ}

u_1=\frac{M+m_1}{m_1}\times \sqrt{2gL(1-\cos 6.8)}

for second case \theta =11.4^{\circ}

u_2=\frac{M+m_2}{m_2}\times \sqrt{2gL(1-\cos 11.4)}

Therefore \frac{u_1}{u_2}=\frac{\frac{M+m_1}{m_1}\times \sqrt{2gL(1-\cos 6.8)}}{\frac{M+m_2}{m_2}\times \sqrt{2gL(1-\cos 11.4)}}

\frac{u_1}{u_2}=\frac{1819.181\times 0.0838}{1001\times 0.1404}

\frac{u_1}{u_2}=1.084

i.e.\frac{v_1}{v_2}=1.084

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Anon25 [30]

Answer:

0

Explanation:

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