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2 years ago

A pan-galactic gargle (an alien) whose mass is 85 kg exploring a large asteroid. What is the force of

Physics

1 answer:

rosijanka [135]2 years ago

7 0

The acceleration due to gravity of the asteroid is 833.85 N.

<h3>Force of gravity holding the the alien down</h3>

F = mg

where;

m is mass of the alien
g is acceleration due to gravity

Assuming the acceleration due to gravity of the asteroid = 9.81 m/s²

F = 85 x 9.81

F = 833.85 N

Thus, the acceleration due to gravity of the asteroid is 833.85 N.

Learn more about force of gravity here: brainly.com/question/2537310

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You might be interested in

A student pushes a 2.85 kg cart causing it to accelerate at a rate of 4.9 m/s squared .What amount of force must the student hav

uysha [10]

In order to find the force (F), you would have to use the formula for it:
F=ma
where m is mass and a is acceleration.
In the problem, the mass is 2.85kg and the acceleration is 4.9m/s^2.
Therefore,
F=2.85kg(4.9m/s^2)
F=13.965kg(m/s^2)
Since N=kg(m/s^2)
F=13.965N
And because the problem requires that we use only 2 significant figures,
F=13N
Therefore, the student must exert 13N of force.

7 0

4 years ago

A bus with a vertical wind shield moves horizontally

Katena32 [7]

Answer:

(3) 53°

Explanation:

We want to measure the angle that the rains form with the vertical wind shield, so we have to measure the angle relative to the vertical. This means that we can write the following equation

$tan \theta = \frac{v_x}{v_y}$

where

$v_y = 30 km/h$ is the speed of the rain, which travels vertically

$v_x = 40 km/h$ is the speed of the bus, which travels horizontally

Substituting, we find

$\theta=tan^{-1}(\frac{40}{30})=53^{\circ}$

8 0

3 years ago

How does a concave mirror form an image

vladimir2022 [97]

Answer:

The correct option is (c) "it bounces the light towards a focal point"

Explanation:

There are two types of spherical mirrors i.e. concave and convex.

A concave mirror is a type of spherical mirror that is curved inwards. For this type of mirror the rays of light are bounced back and converge to the focal point of the mirror.

So, the correct option is (c).

8 0

3 years ago

Three cars (car F, car G, and car H) are moving with the same speed and slam on their brakes. The most massive car is car F, and

Crazy boy [7]

To solve this problem it is necessary to apply the concepts related to Normal Force, frictional force, kinematic equations of motion and Newton's second law.

From the kinematic equations of motion we know that the relationship of acceleration, velocity and distance is given by

$v_f^2=v_i^2+2ax$

Where,

$v_f =$ Final velocity

$v_i =$ Initial Velocity

a = Acceleration

x = Displacement

Acceleration can be expressed in terms of the drag coefficient by means of

$F_f = \mu_k (mg) \rightarrow$ Frictional Force

$F = ma \rightarrow$ Force by Newton's second Law

Where,

m = mass

a= acceleration

$\mu_k =$ Kinetic frictional coefficient

g = Gravity

Equating both equation we have that

$F_f = F$

$\mu_k mg=ma$

$a = \mu_k g$

Therefore,

$v_f^2=v_i^2+2ax$

$0=v_i^2+2(\mu_k g)x$

Re-arrange to find x,

$x = \frac{v_i^2}{2(-\mu_k g)}$

The distance traveled by the car depends on the coefficient of kinetic friction, acceleration due to gravity and initial velocity, therefore the three cars will stop at the same distance.

3 0

3 years ago

A net force of –8750N is used to stop of 1250.kg car travelling 25m/s. What braking distance is needed to bring the car to a hal

Karolina [17]

Answer:

d = 44.64 m

Explanation:

Given that,

Net force acting on the car, F = -8750 N

The mass of the car, m = 1250 kg

Initial speed of the car, u = 25 m/s

Final speed, v = 0 (it stops)

The formula for the net force is :

F = ma

a is acceleration of the car

$a=\dfrac{F}{m}\\\\a=\dfrac{-8750}{1250}\\\\a=-7\ m/s^2$

Let d be the breaking distance. It can be calculated using third equation of motion as :

$v^2-u^2=2ad\\\\d=\dfrac{v^2-u^2}{2a}\\\\d=\dfrac{0^2-(25)^2}{2\times (-7)}\\\\d=44.64\ m$

So, the required distance covered by the car is 44.64 m.

4 0

3 years ago