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3 years ago

How can a heavy moving van have the same momentum as a small motorcycle?

1 answer:

vladimir1956 [14]3 years ago

3 0

Hello there!

The formula for momentum is $p=mv$ .
p is momentum, m is mass, and v is velocity.

The mass of a heavy moving van will be much greater than the mass of a motorcycle, so how can they have the same momentum?

They can have the same momentum if the velocity of the moving van is less than the velocity of the motorcycle, or if the velocity of the motorcycle is greater than the moving van (same thing but differently worded)

Let the mass of the moving van be greater than the mass of the motorcycle by a factor of k. In order for the two to have the same momentum, the velocity of the motorcycle must be greater than the velocity of the moving van by the same factor, k.

Hope this helps! :)

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Which is not expecting acceleration?

Elena L [17]

A point on the outside of a spinning top whose rotational speed is constant.

Since the point at the top is at rest so it will not accelerate at all

A skydiver whose air resistance is equal to that of her weight.

Since weight is counter balanced by air resistance so net force on it is zero so acceleration will be zero

A car on the freeway experiencing a net force of -120 N.

Since net force is not zero so it will accelerate here

A submerged beach ball whose buoyant force is eight times the force of gravity on it.

Since buoyant force is more than the weight so it will not at rest and move upwards with some acceleration

A frictionless spinning merry go round.

Since spinning wheel accelerating due to centripetal force so it is non inertial frame

A falling rock.

Falling rock is accelerating due to gravity so it is again non inertial frame

A hot air balloon moving at 30 degrees east of north with no net force.

since hot air balloon is having no force on it so its acceleration is zero and its an inertial frame

A space shuttle whose boosters just ignited for takeoff.

Since space shuttle is accelerating due to boosters so it is non inertial frame

Here net force on it is due to two given forces

$F_1 = 4320 N$ towards right

$F_2 = 4380 N$ towards left

so here net force is given as

$F = F_2 - F_1$

$F = 4380 - 4320 = 60 N$

since force towards left is more so the direction will be towards left

so correct answer is 60 N towards left

4 0

3 years ago

A girl stands on the edge of a merry-go-round of radius 1.71 m. If the merry go round uniformly accerlerates from rest to 20 rpm

Mashutka [201]

Answer:

$a = 0.53 m/s^2$

Explanation:

initially the merry go round is at rest

after 6.73 s the merry go round will accelerates to 20 rpm

so final angular speed is given as

$\omega = 2\pi f$

$\omega = 2\pi ( \frac{20}{60})$

$\omega = 2.10 rad/s$

so final tangential speed is given as

$v = r\omega$

$v = 1.71 (2.10) = 3.58 m/s$

now average acceleration of the girl is given as

$a = \frac{v_f - v_i}{\Delta t}$

$a = \frac{3.58 - 0}{6.73}$

$a = 0.53 m/s^2$

8 0

3 years ago

Which of the following best describes the circuit shown below?

Rudiy27

Answer:

Short circuit

Explanation:

The given figure shows a short circuit. It is defined as the circuit which allows the flow of electric current when there is no resistance. It shows a battery, bulb and connecting wires.

The wire across the bulb is connected from one terminal to another without any resistance in between them.

So, the correct option is (d) " short circuit ".

3 0

3 years ago

Read 2 more answers

Convert 12cm to picometers

Salsk061 [2.6K]

Answer:

1.2e+11

Explanation:

8 0

3 years ago

Read 2 more answers

Hello guys! Can u please help me with physics. I translated it in English. Can yall help me please how much u can!!

DedPeter [7]

1. Since the body is thrown vertically upward, the only force acting on it as it rises and falls is gravity, which causes a constant downward acceleration with magnitude g = 9.8 m/s². Because this acceleration is constant, we can use the formula

v² - u² = 2a ∆x

where

u = initial speed

v = final speed

a = acceleration

∆x = displacement

At its maximum height, some distance y above the point where the body is launched, the body has zero velocity, so

0² - (20 m/s)² = 2 (-9.8 m/s²) y

Solve for y :

y = (20 m/s)² / (2 (9.8 m/s²)) ≈ 20.4 m

2. Relative to the ground, the body's maximum height is 60 m + 20.4 m ≈ 80.4 m.

3. At any time t ≥ 0, the body's vertical velocity is given by

v = 20 m/s - gt

At the highest point, we have

0 = 20 m/s - (9.8 m/s²) t

and solving for t gives

t = (20 m/s) / (9.8 m/s²) ≈ 2.04 s

4. The body's height y above the ground at any time t ≥ 0 is given by

y = 60 m + (20 m/s) t - 1/2 gt²

Solve for t when y = 0 :

0 = 60 m + (20 m/s) t - 1/2 (9.8 m/s²) t²

Using the quadratic formula,

t = (-b + √(b² - 4ac)) / (2a)

(and omitting the negative root, which gives a negative solution) where a = -1/2 (9.8 m/s²), b = 20 m/s, and c = 60 m. You should end up with

t ≈ 6.09 s

5. At the time found in (4), the body's velocity is

v = 20 m/s - g (6.09 s) ≈ -39.7 m/s

Speed is the magnitude of velocity, so the speed in question is 39.7 m/s.

6 0

3 years ago