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Alja [10]
4 years ago
14

An indication of how close a measurement is to the correct result

Physics
1 answer:
mezya [45]4 years ago
3 0
Accuracy is the correct answer. I remember reading about it.
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135,000 kilometers = how many miles?
DerKrebs [107]
1 mile = 1.609 km

(135,000 km) x (1 mile / 1.609 km)  =  83,885.1 miles
5 0
3 years ago
What is the difference between revolution and rotation
Sphinxa [80]
A rotation is when the earth turns in a circle on it's axis. A revolution is the oblong path the the earth is constantly moving in around the sun. A revolution takes 365.25 days where as a rotation happens in 24 hours. <span />
8 0
3 years ago
Read 2 more answers
What distance will be traveled if you are going 120km/hr for 0.75 hr?
SVEN [57.7K]

Answer:

60 km

Explanation:

For an object (or a person, such as in this case) moving at constant speed, the speed is equal to the ratio between the distance travelled and the time taken:

where

v is the speed

d is the distance

t is the time taken

In this case, we have:

v = 120 km/h is the speed

t = 30 min = 0.5 h is the time taken

Therefore, we can rearrange the equation to find the total distance travelled:

5 0
3 years ago
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

4 0
3 years ago
The flow rate in a firehose is 0.524 m3/s. It is able to shoot water to the top of a building 40.4 m tall, but not higher. You r
Marrrta [24]

Answer:

A fire hose must be able to shoot water to the top of a building 35.0 m tall ... Water enters this hose at a steady rate of 0.500 m3/s and shoots out of a round nozzle. ... I know that Flow rate=0.500 m3/s=A*V. I know the pressure needed to ... The first equation has no potential while the second has no kinetic.

Explanation:

3 0
3 years ago
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