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2 years ago

A force of 50 Newtons causes an object to accelerate at 10 meters per second squared. What is the mass of the object?

Physics

1 answer:

babymother [125]2 years ago

4 0

Use F = ma, where F is force (in newtons), m is mass (in kg) and a is acceleration (in m/s²).

In this case F = 50 N and a = 10 m/s², so we have the equation:

50 = 10m.

Now divide both sides by 10, getting:

m = 5 kg.

So the answer is C. 5.0kg

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An astronaut is in equilibrium when he is positioned 140 km from the center of asteroid C and 581 km from the center of asteroid

tekilochka [14]

Answer:B

Explanation:

Given

Distance of astronaut From asteroid x is $r_x=140 km$

Distance of astronaut From asteroid Y is $r_y=581 km$

Suppose M,M_x,M_y be the masses of Astronaut , asteroid X and Y

If the astronaut is in equilibrium then net gravitational force on it is zero

$F_x=F_y$

$\frac{GMM_x}{r_x^2}=\frac{GMM_y}{r_y^2}$

cancel out the common terms we get

$\frac{M_x}{r_x^2}=\frac{M_y}{r_y^2}$

$\frac{M_x}{M_y}=(\frac{r_x}{r_y})^2$

$\frac{M_x}{M_y}=(\frac{140}{581})^2$

$\frac{M_x}{M_y}=0.05806\approx 0.0581$

4 0

2 years ago

Please help! will give brainlest!!!!!!!!!!!!

eimsori [14]

The force of friction is 19.1 N

Explanation:

According to Newton's second law, the net force acting on the bag is equal to the product between its mass and its acceleration:

$\sum F = ma$

where

$\sum F$ is the net force

m is the mass

a is the acceleration

The bag is moving at constant speed, so its acceleration is zero:

$a=0$

Therefore the net force is zero as well:

$\sum F = 0$

Here we are interested only in the forces acting along the horizontal direction, therefore the net force is given by:

$\sum F = F cos \theta - F_f = 0$

where

$F cos \theta$ is the horizontal component of the applied force, with

F = 22.5 N

$\theta=32.0^{\circ}$

$F_f$ is the force of friction

And solving for $F_f$ , we find

$F_f =Fcos \theta=(22.5)(cos 32.0^{\circ})=19.1 N$

Learn more about friction:

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#LearnwithBrainly

7 0

2 years ago

block of mass 0.5kg on a horizontal surface is attached to a horizontal spring of negligible mass and spring constant 50N/m . Th

Alisiya [41]

Answer:

Explanation:

The mass of the block is 0.5kg

m = 0.5kg.

The spring constant is 50N/m

k =50N/m.

When the spring is stretch to 0.3m

e=0.3m

The spring oscillates from -0.3 to 0.3m

Therefore, amplitude is A=0.3m

Magnitude of acceleration and the direction of the force

The angular frequency (ω) is given as

ω = √(k/m)

ω = √(50/0.5)

ω = √100

ω = 10rad/s

The acceleration of a SHM is given as

a = -ω²A

a = -10²×0.3

a = -30m/s²

Since we need the magnitude of the acceleration,

Then, a = 30m/s²

To know the direction of net force let apply newtons second law

ΣFnet = ma

Fnet = 0.5 × -30

Fnet = -15N

Fnet = -15•i N

The net force is directed to the negative direction of the x -axis

8 0

2 years ago

Tom’s company has been contracted to excavate uranium ore with minimal ground disruption. What process should his company use?

ziro4ka [17]

In-situ leaching or solution mining offers the least ground disruptive type of mining and waste. This type of mining only dissolves the uranium where it is under the ground then pump up to the ground and further processed through milling.

7 0

2 years ago

Read 2 more answers

A box is being pulled by two ropes. Eduardo pulls to the left with a force of 500 N, and Clara pulls to the right with a force o

sergejj [24]

Answer:

kinetic friction or something else

Explanation:

Draw a diagram to illustrate the problem, as shown below.

m = mass of the box, kg

mg = weight of the box, N, acting downward

Because of Newton's 3rd law, a normal force of N = mg acts upward on the box.

C = 200 N, the force applied by Clara, horizontally to the right

E = 500 N, the force applied by Eduardo, horizontally to the left.

Because E is greater than C, the box will move left.

The frictional force, F, will resist the initial motion. Its magnitude is

F = μN = μmg, where μ = kinetic coefficient of friction.

F acts horizontally to the right.

From the given table, we can conclude that

Normal force acting upward is correct

Tension by Eduardoacting left is correct.

Tension by Clara acting right is correct.

Kinetic friction acting left is Incorrect.

Gravity acting downward is correct.correct

Answer:

The direction of the kinetic friction force is incorrect

8 0

2 years ago