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3 years ago

12

Mr. Hoffman, a science teacher, drove 10 miles to school from home in 20 minutes. He drove the 10 miles home in 30 min. His aver

age velocity was _____ m/s.

2 answers:

olga2289 [7]3 years ago

7 0

His average velocity for the day was 0. He ended where he started so his displacement was 0.

Andrews [41]3 years ago

4 0

zero.

Because he finished right where he started. his total displacement zero. therefore, his average velocity was also zero

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008 (part 3 of 4) 3.0 points

motikmotik

Car A take a time of 2.55hr and car B take a time of 2.14 hr

We know that distance divide by time is speed

here it is given that car A to reach a gas station a distance 189 km from the school traveling at a speed of 74 km/hr

so speed=distance/time

s=d/t

t=d/s

=189/74

=2.55hr

In case of car B it is given that The distance from the is 199.8km, car b is traveling at a speed of 93 km/hr

s=d/t

t=d/s

=199.8/93

=2.14hr

so from the above given data and the formula we solved and found out the time taken by car A is 2.55h and car B is 2.14h

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11 months ago

A car accelerates from rest to a velocity of 5 meters/second in 4 seconds. What is its average acceleration over this period of

disa [49]

The average acceleration is

$\bar a=\dfrac{5\,\frac{\mathrm m}{\mathrm s}-0\,\frac{\mathrm m}{\mathrm s}}{4\,\mathrm s}=1.25\,\dfrac{\mathrm m}{\mathrm s^2}$

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3 years ago

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A ball is thrown vertically upward. What is its acceleration right before it hits the earth?

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3 years ago

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Explain what it means when we say a substance is magnetic

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3 years ago

A 460 g model rocket is on a cart that is rolling to the right at a speed of 3.0 m/s. The rocket engine, when it is fired, exert

ollegr [7]

Answer:

The rocket should be launched at a horizontal distance of <u>6.45 m</u> left of the loop.

Explanation:

Given:

Mass of the rocket model (m) = 460 g = 0.460 kg [1 g = 0.001 kg]

Speed of the cart (v) = 3.0 m/s

Thrust force by the rocket engine (F) = 8.5 N

Vertical height of the loop (y) = 20 m

Let the horizontal distance left to the loop for launch be 'x'. Also, let 't' be the time taken by the rocket to reach the loop.

Now, there are two types of motion associated with the rocket- one is horizontal and the other vertical.

So, we will apply kinematics of motion in the two directions separately.

Vertical motion:

Given:

Force acting in the vertical direction is given as:

$F_y=F-mg=8.5-0.46\times 9.8=3.992\ N$

So, acceleration in the vertical direction is given as:

Acceleration = Force ÷ mass

$a_y=\frac{F_y}{m}=\frac{3.992\ N}{0.46\ kg}=8.678\ m/s^2$

Vertical displacement of rocket is same as the height of loop. So, $y=20\ m$

There is no initial velocity in the vertical direction. So, $u_y=0\ m/s$

Now, applying equation of motion in vertical direction. we have:

$y=u_yt+\frac{1}{2}a_yt^2\\\\20=0+\frac{1}{2}\times 8.678t^2\\\\20=4.339t^2\\\\t^2=\frac{20}{4.339}\\\\t=\sqrt{\frac{20}{4.339}}=2.15\ s$

Now, time taken to reach the loop is 2.15 s.

Horizontal motion:

There is no acceleration in the horizontal motion. So, displacement in the horizontal direction is equal to the product of horizontal speed and time.

Also, displacement of the rocket in the horizontal direction is nothing but the horizontal distance of its launch left of the loop. So,

$x=vt\\\\x=3.0\ m/s\times 2.15\ s\\\\x=6.45\ m$

Therefore, the rocket should be launched at a horizontal distance of 6.45 m left of the loop.

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3 years ago