How much work is done when you lift a 5kg bag of groceries 1.5 m?

Suppose an object in freefall is dropped from a building it’s starting velocity is 0m/s. Ignoring the facts of air resistance wh

Alja [10]

v = v₀ + at

v = final speed, v₀ = initial speed, a = acceleration, t = elapsed time

Given values:

v₀ = 0m/s (starts from rest), a = 9.81m/s², t = 3s

Plug in and solve for v:

v = 0 + 9.81(3)

v = 29.4m/s

3 0

3 years ago

1. A rocket is forced forward by the ______ force of its engines, expelling gases out the rear of the rocket.

AURORKA [14]

There are two forces acting on a rocket at the moment of lift off: Thrust pushes the rocket upwards by pushing gases downwards in the opposite direction.Weight is the force due to gravity pulling the rocket downwards towards the centre of the earth.So I'm thinking the answer is THRUST.

6 0

3 years ago

A Boeing 777 aircraft has a mass of 300,000 kg. At a certain instant during its landing, its speed is 27.0 m/s. If the braking f

GarryVolchara [31]

Answer:

Speed of the airplane 10.0 s later = 12.2 m/s

Explanation:

Mass of Boeing 777 aircraft = 300,000 kg

Braking force = 445,000 N

Deceleration

$a=\frac{445000}{300000}=1.48m/s^2$

Initial velocity, u = 27 m/s

Time , t = 10 s

We have equation of motion, v =u +at

v = 27 + (-1.48) x 10 = 27 - 14.8 = 12.2 m/s

Speed of the airplane 10.0 s later = 12.2 m/s

6 0

3 years ago

A container with volume 1.64 L is initially evacuated. Then it is filled with 0.226 g of N2N

vaieri [72.5K]

Answer:

0.015 atm

Explanation:

The pressure of the gas can be calculated using Ideal Gas Law:

$p = \frac{nRT}{V}$

Where:

n: is the number of moles of the gas

R: is the gas constant = 0.082 L*atm/(K*mol)

V: is the volume of the container = 1.64 L

T: is the temperature

We need to find the number of moles and the temperature. The number of moles is:

$n = \frac{m}{M}$

Where:

M: is the molar mass of the N₂ = 14.007 g/mol*2 = 28.014 g/mol

m: is the mass of the gas = 0.226 g

$n = \frac{0.226 g}{28.014 g/mol} = 8.07 \cdot 10^{-3} moles$

Now, the temperature can be found using the following equation:

$v_{rms} = \sqrt{\frac{3RT}{M}}$

Where:

R: is the gas constant = 0.082 L*atm/K*mol = 8.314 J/K*mol

$v_{rms}$ : is the root-mean-square speed of the gas = 182 m/s

By solving the above equation for T, we have:

$T = \frac{v_{rms}^{2}*M}{3R} = \frac{(182 m/s)^{2}*28.014 \cdot 10^{-3} Kg/mol}{3*8.314 J K^{-1}mol^{-1}} = 37.20 K$

Finally, we can find the pressure of the gas:

$p = \frac{nRT}{V} = \frac{8.07 \cdot 10^{-3} mol*0.082 L*atm* K^{-1}*mol^{-1}*37.20 K}{1.64 L} = 0.015 atm$

Therefore, the pressure of the gas is 0.015 atm.

I hope it helps you!

8 0

3 years ago

A 1000-kg car is driving toward the north along a straight horizontal road at a speed of 20.0 m/s. The driver applies the brakes

Nitella [24]

Answer:

The value of F= - 830 N

Since the force is negative, it implies direction of the force applied was due south.

Explanation:

Given data:

Mass = 1000-kg

Distance, d = 240 m

Initial velocity, v1 = 20.0 m/s

Final velocity, v2 = 0 (since the car came to rest after brake was applied)

v2²= v1² + 2ad (using one of the equation of motion)

0= 20² + (2 x a x 240)

0= 400 + 480 a

a = - 400/480

a = - 0.83 m/s²

Then, imputing the value of a into

F = ma

F = 1000 kg x ( - 0.83 m/s²)

F= - 830 N

The car was driving toward the north, and since the force is negative, it implies direction of the force applied was due south.

3 0

3 years ago