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3 years ago

7

Megan rode the bus to school, which is located 8 kilometers from her home. If Megan's frame of reference is her house, and it to

ok her 15 minutes to get home, what was the bus' average velocity during the return trip?

1 answer:

Dmitriy789 [7]3 years ago

7 0

Answer:

Explanation: idk sry

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A race car starts from rest and travels east along a straight and level track. For the first 5.0 s of the car’s motion, the east

Marrrta [24]

Answer:

ax = 6.43m/s²

Explanation:

The acceleration is the time derivative of the velocity function ax = dvx(t)/dt

We have been given the velocity function v(t) and also the velocity v = 12.0m/s and we are requested to calculate the acceleration at this time which we don't know.

So the first step is to calculate the time at which the velocity =12.0m/s and with this time calculate the acceleration. Detailed solution can be found in the attachment below.

7 0

3 years ago

An object with total mass mtotal = 14.6 kg is sitting at rest when it explodes into three pieces. One piece with mass m1 = 4.9 k

zheka24 [161]

Answer: 1) 0. 2) 4.2 Kg. 3) 15.4 m/s 4) 12.9 m/s 5) 0. 6) 3.62 KJ.

Explanation:

1) Assuming that no external forces act during the collision, total momentum must be conserved. As initially the total mass was at rest, so initial momentum is zero, final momentum of all the system must be 0 also.

2) After the explosion, as mass must be conserved also, the sum of the masses of the three pieces must be equal to the original total mass, so we can write the following:

m₁ + m₂ + m₃ = M = 14.6 Kg = 4.9 Kg + 5.5 Kg + m₃

Solving for m₃, we have:

m₃ = 14.6 Kg - 4.9 Kg -5.5 Kg = 4.2 Kg.

3) and 4)

As momentum is a vector, if it is magnitude must be 0, this means that all his components must be 0 too.

So, we can write two equations, one for the x-component, and other for the y-component, as follows:

pₓ = m₁. v₁ₓ + m₂.v₂ₓ + m₃.v₃ₓ = 0

py = m₁.v₁y + m₂. v₂y + m₃. v₃y =0

Replacing by the values, and solving for v₃ₓ and v₃y, we get:

v₃ₓ = 15.4 m/s

v₃y = 12.9 m/s

v = √(15.4)²+(12.9)² = 20.1 m/s

5) As the center of mass must move as if all the mass were concentrated in this point, and we know that the total momentum must be 0, this tells us that the magnitude of the velocity of the center of mass must be 0 too.

6) As initial kinetic energy is 0, as the mass was at rest, the increase in the kinetic energy is obtained simply adding the kinetic energy of every piece of mass gained after explosion, as follows:

K = K₁ + K₂ + K₃ = 1/2 (m₁ . v₁² + m₂.v₂² + m₃.v₃²)

Replacing by the values, we get:

K= 3.62 KJ

4 0

3 years ago

Two friends are playing a version of proton golf where the hole is marked by a single proton. The first friend reads his meter,

yarga [219]

Answer:

the electric field strength on the second one is 2.67 N/C.

Explanation:

the electric fiel on the first one is:

E1 = k×q/(r^2)

r^2 = k×q/(E1)

= (9×10^9)×(q)/(24.0)

= 375000000q

then the electric field on the second one is:

E2 = k×q/(R^2)

we know that R = 3r

R^2 = 9×r^2

E2 = k×q/(9×r^2)

= k×q/(9×375000000q)

= k/(9×375000000)

= (9×10^9)/(9×375000000)

= 2.67 N/C

Therefore, the electric field strength on the second one is 2.67 N/C.

5 0

3 years ago

A rocket ship has several engines and thrusters. While the Solid Rocket Booster (SRB) and main engines only work together during

Katarina [22]

A rocket ship is accelerated by the SRB and the main engines for 2.0 minutes and the main engines for 8.5 minutes after the launch. The acceleration of the ship during the first 2.0 minutes is 11 m/s² (D).

A rocket ship has several engines and thrusters. We can divide its initial movement into 2 parts:

From t = 0 min to t = 2.0 min, the SRB and the main engines act together and the speed goes from 0 m/s (rest) to 1341 m/s.
From t = 2.0 min to t = 8.5 min, the main engines alone accelerate the ship form 1341 m/s to 7600 m/s.

We want to know the acceleration in the first part (first 2.0 minutes). We need to consider that:

The speed increases from 0 m/s to 1341 m/s.
The time elpased is 2.0 min.
1 min = 60 s.

The acceleration of the ship during the first 2.0 minutes is:

$a = \frac{\Delta v }{t} ) \frac{(1341m/s-0m/s)}{2.0min} \times \frac{1min}{60s} = 11 m/s^{2}$

A rocket ship is accelerated by the SRB and the main engines for 2.0 minutes and the main engines for 8.5 minutes after the launch. The acceleration of the ship during the first 2.0 minutes is 11 m/s² (D).

Learn more: brainly.com/question/16274121

3 0

2 years ago

In the metric system, the appropriate unit for weight is the _____. gram newton newton/cm2 gram/cm3

Archy [21]

Answer:

Newton

Explanation:

The earth attracts every body towards its centre. The force with which the earth attracts any body towards its centre, is called its weight.

It is a vector quantity.

It always acts towards the centre of earth.

The SI unit of Newton.

4 0

3 years ago