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Alex73 [517]
2 years ago
15

What is the weight of an object (mass = 60 kilograms) on Mars, where the acceleration due to gravity is 3.75 meters/second2?. Se

lect one of the options below as your answer:. A. 1.6 newtons. B. 16 newtons. C. 22.5 newtons. D. 225 newtons. E. 2250 newtons.
Physics
1 answer:
german2 years ago
6 0
Weight = mass * gravity = 60 kg * 3.75 m/s² = 225 N

<span>Option D.</span>
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A scuba diver has his lungs filled to half capacity (3 liters) when 10 m below the surface. If the diver holds his breath while
rjkz [21]

To solve this problem it is necessary to apply Boyle's law in which it is specified that

P_1V_1 =P_2 V_2

Where,

P_1 and V_1 are the initial pressure and volume values

P_2 and V_2 are the final pressure volume values

The final pressure here is the atmosphere, then

P_2 = 101325 \approx 1*10^5Pa

h = 10m

\rho_w = 1000kg/m^3

V_1 = 3.0L

Pressure at the water is given by,

P_1 = P_2 -\rho gh

P_1 = 1*10^5 +1000*9.8*10 =198000Pa

Using Boyle equation we have,

V_2 = \frac{P_1V_1}{P_2}

V_2 = \frac{198000*3*10^5}{10^5}

V_2 = 5.9L

Therefore the volume of the lungs at the surface is 5.9L

4 0
2 years ago
A block of mass 3m is placed on a frictionless horizontal surface, and a second block of mass m is placed on top of the first bl
tatuchka [14]

By Newton's second law, assuming <em>F</em> is horizontal,

• the net <u>horizontal</u> force on the <u>larger</u> block is

<em>F</em> - <em>µmg</em> = 3<em>mA</em>

where <em>µmg</em> is the magnitude of friction felt by the larger block due to rubbing with the smaller one, <em>µ</em> is the coefficient of static friction between the two blocks, and <em>A</em> is the block's acceleration;

• the net <u>vertical</u> force on the <u>larger</u> block is

4<em>mg</em> - 3<em>mg</em> - <em>mg</em> = 0

where 4<em>mg</em> is the mag. of the normal force of the surface pushing up on the combined mass of the two blocks, 3<em>mg</em> is the weight of the larger block, and <em>mg</em> is the weight of the smaller block;

• the net <u>horizontal</u> force on the <u>smaller</u> block is

<em>µmg</em> = <em>ma</em>

where <em>µmg</em> is again the friction between the two blocks, but notice that this points in the same direction as <em>F</em>. It is the only force acting on the smaller block in the horizontal direction, so (b) static friction is causing the smaller block to accelerate;

• the net <u>vertical</u> force on the <u>smaller</u> block is

<em>mg</em> - <em>mg</em> = 0

where <em>mg</em> is the magnitude of both the normal force of the larger block pushing up on the smaller one, and the weight of the smaller block.

(You should be able to draw your own FBD's based on the forces mentioned above.)

(c) Solve the equations above for <em>A</em> and <em>a</em> :

<em>A</em> = (<em>F</em> - <em>µmg</em>) / (3<em>m</em>)

<em>a</em> = <em>µg</em>

5 0
1 year ago
You throw a Frisbee of mass m and radius r so that it is spinning about a horizontal axis perpendicular to the plane of the Fris
Usimov [2.4K]

Answer:

Torque τ =w ×0 = 0

Explanation:

We know that the torque is given by the product of the force and perpendicular distance between the force and the axis.

Here the gravity force act at the center and the rotational axis is also passing through the center.

Therefore the perpendicular distance between the force and the rotational axis would be zero.

Hence the torque will be

Torque = Force × Perpendicular distance

Torque = mg×0 = 0

Therefore the torque would be zero.

7 0
2 years ago
A 2 kg stone is tied to a 0.5 m string and swung around a circle at a constant angular velocity of 12 rad/s. the angular momentu
Illusion [34]
Starting from the angular velocity, we can calculate the tangential velocity of the stone:
v=\omega r= (12 rad/s)(0.5 m)= 6 m/s

Then we can calculate the angular momentum of the stone about the center of the circle, given by
L=mvr
where
m is the stone mass
v its tangential velocity
r is the radius of the circle, that corresponds to the length of the string.

Substituting the data of the problem, we find
L=(2 kg)(6 m/s)(0.5 m)=6 kg m^2 s^{-1}
3 0
2 years ago
The escape velocity is defined to be the minimum speed with which an object of mass m must move to escape from the gravitational
s344n2d4d5 [400]

Answer:

v = √2G M_{earth} / R

Explanation:

For this problem we use energy conservation, the energy initiated is potential and kinetic and the final energy is only potential (infinite r)

        Eo = K + U = ½ m1 v² - G m1 m2 / r1

        Ef = - G m1 m2 / r2

When the body is at a distance R> Re, for the furthest point (r2) let's call it Rinf

       Eo = Ef

       ½ m1v² - G m1 M_{earth} / R = - G m1 M_{earth} / R

      v² = 2G M_{earth} (1 / R - 1 / Rinf)

If we do Rinf = infinity     1 / Rinf = 0

       v = √2G M_{earth} / R

      Ef = = - G m1 m2 / R

The mechanical energy is conserved  

 

      Em = -G m1  M_{earth} / R

      Em = - G m1  M_{earth} / R

     R = int        ⇒  Em = 0

6 0
1 year ago
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