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2 years ago

Need help 8th grade science test need help

Physics

2 answers:

jolli1 [7]2 years ago

7 0

ANSWER: d, average speed

Liula [17]2 years ago

7 0

Answer:

c or d

Explanation:

hopes this helps :)

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A person drives 70 km/h in 1 hour to the east, then 80 km/h for another hour to the east. What

andre [41]

Answer: The average velocity is 150 km/h

Explanation: 70+80=150

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3 years ago

What is the formula that describes the magnitude of impulse on an object?

vladimir1956 [14]

Answer:

Option C.

Impulse = mass × change in velocity

Explanation:

Impulse is defined by the following the following formula:

Impulse = force (F) × time (t)

Impulse = Ft

From Newton's second law of motion,

Force = change in momentum /time

Cross multiply

Force × time = change in momentum

Recall:

Impulse = Force × time

Thus,

Impulse = change in momentum

Recall:

Momentum = mass x velocity

Momentum = mv

Chang in momentum = mass × change in velocity

Change in momentum = mΔv

Thus,

Impulse = change in momentum

Impulse = mass × change in velocity

8 0

3 years ago

If two point masses 1kg & 4kg are seperated by a distance of 2m. Magnitude of gravitational force exerted by 1kg on 4kg is ?

Aliun [14]

Answer:

F = G Newtons

Explanation:

Given:

Mass of 1st body = $1\:kg$
Mass of 2nd body = $4\:kg$

To Find:

Magnitude of gravitational force

Solution:

Here, we have a formula

$F=\dfrac{G.M_{1}.M_{2}}{r^{2}}$

<u>Substituting the values</u>

$\implies\:\:F = \dfrac{G(1)(4)}{2^{2}}$

$\implies\:\:F = \dfrac{4G}{4}$

$\implies\:\:F = \dfrac{\cancel{4}G}{\cancel{4}}$

$\implies\:\:\red{F = G}$

Know More:

The applied formula for the above solution is

${\boxed{F_{G}=\dfrac{G.M_{1}.M_{2}}{r^{2}}}}$

where,

F $_{G}$ = Gravitational force
G = Gravitational constant
M $_{1}$ = mass of 1st body
M $_{2}$ = mass of 2nd body
r = distance between two bodies

6 0

2 years ago

What is the magnitude of the total displacement of the school bus.

Evgesh-ka [11]

Answer:

400 or 500

Explanation:

6 0

2 years ago

Determine the instantaneous velocity of the car at t = 4.7 s, using time intervals of 0.40 s, 0.20 s, and 0.10 s. (In order to b

weeeeeb [17]

Answer:

4.408 m/s, 4.102 m/s, 4.026 m/s

Explanation:

The question is incomplete. The text of the original question states:

A race car moves such that its position fits the relationship

$x=(4.0 m/s)t + (0.85 m/s^3) t^3$

where x is measured in meters and t in seconds. Determine the instantaneous velocity of the car at t = 4.7 s, using time intervals of 0.40 s, 0.20 s, and 0.10 s.

We can find the instantanoues velocity of the car at any time t by calculating the derivative of the position, so we find:

$v(t) = x'(t) = 4.0 m/s + 3\cdot (0.85 m/s^2) t^2 = 4.0 m/s + (2.55 m/s^2) t^2$

And now we just need to substitute t=0.40 s, 0.20 s, and 0.10 s to find the corresponding instantaneous velocity:

$v(0.40) = 4.0 + 2.55 (0.40)^2 = 4.408 m/s\\v(0.20) = 4.0 + 2.55 (0.20)^2 = 4.102 m/s\\v(0.10) = 4.0 + 2.55 (0.10)^2 = 4.026 m/s$

5 0

3 years ago