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PilotLPTM [1.2K]

2 years ago

6

Two planets have the same surface gravity, but planet B has twice the mass of planet A. If planet A has mass m, what is the mass

of planet B

1 answer:

strojnjashka [21]2 years ago

5 0

If B has twice the mass the answer is 2M

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An ant is crawling along a straight wire, which we shall call the x axis, from A to B to C to D

Inga [223]

A high quality insecticide applied all along the axis and the surrounding area can protect against a recurrence for a substantial period of time.

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3 years ago

You are driving your 1700 kg car at 21 m/s down a hill with a 5.0∘ slope when a deer suddenly jumps out onto the roadway. You sl

podryga [215]

Answer:

d = 27.7 m

Explanation:

Here the car is driving on the inclined plane

So here we can say that work done by the gravity and work done by friction is equal to change in kinetic energy of the system

So here we can write it as

$mgsin\theta \times d - F_f \times d = \frac{1}{2}mv_f^2 - \frac{1}{2}mv_i^2$

now we have

m = 1700 kg

$v_f = 0$

$v_i = 21 m/s$

$F_f = 1.5 \times 10^4 N$

$(1700)(9.81)sin5 (d) - (1.5\times 10^4)d = 0 - \frac{1}{2}(1700)(21^2)$

$1453.5 d - (1.5\times 10^4)d = -374850$

$d = 27.7 m$

6 0

3 years ago

You find yourself at a lever. A runaway trolley approaches five people who are tied to a set of tracks. Pulling the lever will d

Musya8 [376]

It’s my guess but from my opinion i would say yes

6 0

3 years ago

A car collides into a concrete wall going 25.0 m/s . It stops in 0.141 seconds and has a change in momentum of 39,400. What is t

vodka [1.7K]

Answer:

Mass of the car is 1576 kg.

Explanation:

Let the mass of the car be $m$ kg.

Given:

Initial velocity of the car is, $u=25\ m/s$

As the car stops, final velocity of the car is, $v=0\ m/s$

Change in momentum is, $\Delta p=39400$

Now, we know that, momentum is given as the product of mass and velocity.

So, change in momentum is given as:

$\Delta p=m(u-v)\\39400=m(25-0)\\39400=25m\\m=\frac{39400}{25}\\m=1576\ kg$

Therefore, the mass of the car is 1576 kg.

4 0

3 years ago

The flywheel of a steam engine runs with a constant angular speed of 113 $rev/min$. When steam is shut off, the friction of the

Oksana_A [137]

Answer:

α = - 1.883 rev/min²

Explanation:

Given

ωin = 113 rev/min

ωfin = 0 rev/min

t = 1.0 h = 60 min

α = ?

we can use the following equation

ωfin = ωin + α*t ⇒ α = (ωfin - ωin) / t

⇒ α = (0 rev/min - 113 rev/min) / (60 min)

⇒ α = - 1.883 rev/min²

6 0

3 years ago