Please look at the photo and give me the answer :)))

A 2 kg object with a weight of 20 N is being pulled up by a rope with a tension of 12N what is the acceleration of the object

son4ous [18]

Answer:

The object accelerates downward at 4 m/s² since the tension on the rope is less than weight of the object.

Explanation:

Given;

mass of the object, m = 2 kg

weigh of the object, W = 20 N

tension on the rope, T = 12 N

The acceleration of the object is calculated by applying Newton's second law of motion as follows;

T = F + W

T = ma + W

ma = T - W

$a = \frac{T-W}{m} \\\\a = \frac{12 - 20}{2} \\\\a = -4 \ m/s^2$ (the negative sign indicates deceleration of the object)

The object accelerates downward at 4 m/s² since the tension on the rope is less than weight of the object.

7 0

2 years ago

A tank contains gas at 13.0°C pressurized to 10.0 atm. The temperature of the gas is increased to 95.0°C, and half the gas is re

fomenos

Answer:

The pressure of the remaining gas in the tank is 6.4 atm.

Explanation:

Given that,

Temperature T = 13+273=286 K

Pressure = 10.0 atm

We need to calculate the pressure of the remaining gas

Using equation of ideal gas

$PV=nRT$

For a gas

$P_{1}V_{1}=nRT_{1}$

Where, P = pressure

V = volume

T = temperature

Put the value in the equation

$10\times V=nR\times286$ ....(I)

When the temperature of the gas is increased

Then,

$P_{2}V_{2}=\dfrac{n}{2}RT_{2}$ ....(II)

Divided equation (I) by equation (II)

$\dfrac{P_{1}V}{P_{2}V}=\dfrac{nRT_{1}}{\dfrac{n}{2}RT_{2}}$

$\dfrac{10\times V}{P_{2}V}=\dfrac{nR\times286}{\dfrac{n}{2}R368}$

$P_{2}=\dfrac{10\times368}{2\times286}$

$P_{2}= 6.433\ atm$

$P_{2}=6.4\ atm$

Hence, The pressure of the remaining gas in the tank is 6.4 atm.

4 0

2 years ago

A spherical asteroid of average density would have a mass of 8.7×1013kg if its radius were 2.0 km.A)If you and your spacesuit ha

WITCHER [35]

A) 0.189 N

The weight of the person on the asteroid is equal to the gravitational force exerted by the asteroid on the person, at a location on the surface of the asteroid:

$F=\frac{GMm}{R^2}$

where

G is the gravitational constant

8.7×10^13 kg is the mass of the asteroid

m = 130 kg is the mass of the man

R = 2.0 km = 2000 m is the radius of the asteroid

Substituting into the equation, we find

$F=\frac{(6.67\cdot 10^{-11})(8.7\cdot 10^{13} kg)(130 kg)}{(2000 m)^2}0.189 N=$

B) 2.41 m/s

In order to orbit just above the surface of the asteroid (r=R), the centripetal force that keeps the astronaut in orbit must be equal to the gravitational force acting on the astronaut:

$\frac{GMm}{R^2}=\frac{mv^2}{R}$