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3 years ago

How do you work out instantaneous speed?

1 answer:

7 0

To calculate instantaneous speed, we need to divide part of the total distance traveled by time. However, we don't want to use the distance of the entire trip, because that will give us average speed.

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What would happen to the amount of matter on earth if mass were not conserved during changes of state?

attashe74 [19]

earth would be thrown off its balance and nature would be in danger of too many resources and not enough resources

4 0

3 years ago

Read 2 more answers

What is the frequency of this wave 1 2 3 4

viva [34]

The frequency of this wave is 3

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3 years ago

If you can run at a speed of 8 miles per hour and you want to run to the store 16 miles away, how

MArishka [77]

Answer:

16÷8=2

Explanation:

if you run 8 mi an hour than in 16 mi you would have run 2 hours

3 0

3 years ago

For an object with a given mass on Earth, calculate the weight of the object with the mass equal in magnitude to the number repr

leonid [27]

Answer: The weight of the object is 29.4 N

Explanation:

To calculate the weight of the object, we use the equation:

$W=m\times g$

where,

m = mass of the object = 3 kg

g = acceleration due to gravity = $9.8m/s^2$

Putting values in above equation, we get:

$W=3kg\times 9.8m/s^2\\\\W=29.4N$

Hence, the weight of the object is 29.4 N

6 0

3 years ago

A 5kg object moving horizontally at 3m/s collides with a stationary 3kg object. After the collision, the 5kg object is deflected

gavmur [86]

Answer:

The velocity of each ball after the collision are 2.19 m/s and 2.58 m/s.

Explanation:

Given that,

Mass of object = 5 kg

Speed = 3 m/s

Mass of stationary object = 3 kg

Moving object deflected = 30°

Stationary object deflected = 31°

We need to calculate the velocity of each ball after collision

Using conservation of momentum

Along x-axis

$m_{1}u_{1}+m_{2}u_{2}=m_{1}v_{1}\cos\theta+m_{2}v_{2}\cos\theta$

Put the value into the fomrula

$5\times3+0=5\times v_{1}\cos30+3\times v_{2}\cos45$

$15=5v_{1}\times\dfrac{\sqrt{3}}{2}+3v_{2}\times\dfrac{1}{\sqrt{2}}$

$15=\dfrac{5\sqrt{3}}{2}v_{1}+\dfrac{3}{\sqrt{2}}v_{2}$ ....(I)

Along y -axis

$m_{1}u_{1}+m_{2}u_{2}=m_{1}v_{1}\sin\theta+m_{2}v_{2}\sin\theta$

Put the value into the formula

$0+0=5\times v_{1}\sin30-3\times v_{2}\sin45$

$\dfrac{5}{2}v_{1}-\dfrac{3}{\sqrt{2}}v_{2}=0$ ...(II)

From equation (I) and (II)

$v_{1}=\dfrac{15\times2}{5\sqrt{3}+5}$

$v_{1}=2.19\ m/s$

Put the value of v₁ in equation (I)

$\dfrac{5}{2}\times2.19-\dfrac{3}{\sqrt{2}}v_{2}=0$

$v_{2}=\dfrac{5.475\times\sqrt{2}}{3}$

$v_{2}=2.58\ m/s$

Hence, The velocity of each ball after the collision are 2.19 m/s and 2.58 m/s.

3 0

3 years ago