All categories

3 years ago

PLEASE HELP ):

Physics

1 answer:

jolli1 [7]3 years ago

6 0

The first thing we must do for this case is the sum of forces in a horizontal direction.

We have then:

$F1 + F2 = m * a$

Substituting values we have:

$50 + 75 = m * 2.5$

From here, we clear the mass of the object:

$m * 2.5 = 125\\m = 125 / 2.5\\m = 50 Kg$

We now look for the weight of the object.

$W = m * g$

Where,

g: acceleration of gravity (9.8 m/s^2)

Substituting values:

$W = 50 * 9.8\\W = 490 N$

Answer:

the weight of the object is:

$W = 490 N$

option 4

You might be interested in

A 2.00-kg object A is connected with a massless string across a massless, frictionless pulley to a 3.00-kg object B. Object A re

slamgirl [31]

Answer:

tension: 19.3 N
acceleration: 3.36 m/s^2

Explanation:

Given

mass A = 2.0 kg

mass B = 3.0 kg

θ = 40°

Find

The tension in the string

The acceleration of the masses

Solution

Mass A is being pulled down the inclined plane by a force due to gravity of ...

F = mg·sin(θ) = (2 kg)(9.8 m/s^2)(0.642788) = 12.5986 N

Mass B is being pulled downward by gravity with a force of ...

F = mg = (3 kg)(9.8 m/s^2) = 29.4 N

The tension in the string, T, is such that the net force on each mass results in the same acceleration:

F/m = a = F/m

(T -12.59806 N)/(2 kg) = (29.4 N -T) N/(3 kg)

T = (2(29.4) +3(12.5986))/5 = 19.3192 N

Then the acceleration of B is ...

a = F/m = (29.4 -19.3192) N/(3 kg) = 3.36027 m/s^2

The string tension is about 19.3 N; the acceleration of the masses is about 3.36 m/s^2.

3 0

3 years ago

The average distance from earth to the sun is 150 megameters. what is that distance in meters?

alexira [117]

It is 150,000,000 meters!!

7 0

3 years ago

A woman of mass 50 kg is swimming with a velocity of 1.6 m/s. If she stops stroking and glides to a stop in the water, what is t

Snezhnost [94]

Answer:

Impulse of force = -80 Ns

Explanation:

Given the following data;

Mass = 50kg

Initial velocity = 1.6m/s

Since she glides to a stop, her final velocity equals to zero (0).

Now, we would find the change in velocity.

$Change \; in \; velocity = final \; velocity - initial \; velocity$

Substituting into the equation above;

Change in velocity = 0 - 1.6 = 1.6m/s

$Impulse \; of \; force = mass * change \; in \; velocity$

Substituting into the equation, we have;

$Impulse \; of \; force = 50 * -1.6$

Impulse of force = -80 Ns

Therefore, the impulse of the force that stops her is -80 Newton-seconds and it has a negative value because it is working in an opposite direction, thus, bringing her to a stop.

5 0

2 years ago

For this problem, we assume that we are on planet-i. the radius of this planet is r =4200 km, the gravitational acceleration at

Minchanka [31]

The expression commonly used for potential gravitational energy is just simplification. It is actually just the first term in Taylor expansion of the real expression.
In general, the potential energy of gravitational field is defined as:
$U=-G \frac{mM}{r}$
Where G is universal gravitational constant, and r is the distance between the objects centers of mass. Negative sign represents the bound state.
Since we are not given the mass of the planet we have to calculate it.
$F_g=G\frac{mM}{r_p^2}\\ mg=G\frac{mM}{r_p^2}\\ g=G\frac{M}{r_p^2}$
This formula can be used for any planet. It gives you the gravitational acceleration on the planet's surface. We can use it to calculate the planet's mass:
$g=G\frac{M}{r_p^2}\\ M=\frac{gr_p^2}{G}=2.41\cdot 10^{24}kg$
Now we can calculate the potential energy of that cannonball when it reaches its maximum height.
$U=-G \frac{mM}{r}\\ U=-G \frac{mM}{r_p+h}$
When we plug in the numbers we get:
$U=-4.99\cdot 10^{10} J$
The potential energy has to be equal to the kinetic energy.
$E_k=4.99\cdot 10^{10} J$

3 0

3 years ago

A sinusoidal electromagnetic wave from a radio station passes perpendicularly through an open window that has area 0.320 m2. at

AlexFokin [52]

Remember, half of the energy in an EM wave is in the E field, the rest is in the B field. Thus, multiply E field energy by 2.
To calculate the energy of the wave you must then use the following equation: W = A*t*c*2*(1/2*E^2*Eo). Where, A = Area, t = time, c = speed of light (which is a constant), E = Electric field, E0 = vacuum permittivity (8.85*10^-12 Nm^2/C^2). Substituting W =(0.320)*(26)*(3*10^8)*(2)*((1/2)*(1.95*10^-2)^2*(8.854*10^-12)) = 8.40*10^-6 J

5 0

3 years ago