All categories

3 years ago

Please help me! I don't understand how to do this!

Physics

1 answer:

dezoksy [38]3 years ago

3 0

From the information in the question, the acceleration of the cart is 2.55 ms-2.

We can find the acceleration using the following information from the question;

u = 0.9 m/s

v = 6 m/s

a = 2.0 seconds

But;

a = v - u/t

v = final velocity

u = initial velocity

t = time

a = 6 - 0.9/2 = 2.55 ms-2

Force= ma

m = mass

a = acceleration

F = 6 kg × 2.55 ms-2

F = 15.3 N

Learn more about acceleration: brainly.com/question/12134554

You might be interested in

A car is moving 5.82M/S when it accelerates at 2.35M/S^2for 3.25S. What is its final velocity?

rodikova [14]

Answer:

Explanation:

vf=vi+at

vi=5.82 m/s

a=2.35 m/s2

t=3.25 s

vf=5.82+2.35*3.25

vf=5.82+7.64

vf=13.46

3 0

3 years ago

An object has a mass of 10.2 kg, what is its weight in Newton's?

Juliette [100K]

Answer:

The answer is 98 N

Explanation:

I am pretty sure

3 0

3 years ago

Read 2 more answers

2. An athlete of average size is hanging from the end of a 20 m long rope, which has a mass of 4 kg and is attached to a hook in

a_sh-v [17]

Answer:

t = 0.319 s

Explanation:

With the sudden movement of the athlete a pulse is formed that takes time to move along the rope, the speed of the rope is given by

v = √T/λ

Linear density is

λ = m / L

λ = 4/20

λ = 0.2 kg / m

The tension in the rope is equal to the athlete's weight, suppose it has a mass of m = 80 kg

T = W = mg

T = 80 9.8

T = 784 N

The pulse rate is

v = √(784 / 0.2)

v = 62.6 m / s

The time it takes to reach the hook can be searched with kinematics

v = x / t

t = x / v

t = 20 / 62.6

t = 0.319 s

7 0

3 years ago

Say that you are in a large room at temperature TC = 300 K. Someone gives you a pot of hot soup at a temperature of TH = 340 K.

DiKsa [7]

Answer:0.061

Explanation:

Given

$T_C=300 k$

Temperature of soup $T_H=340 K$

heat capacity of soup $c_v=33 J/K$

Here Temperature of soup is constantly decreasing

suppose T is the temperature of soup at any instant

efficiency is given by

$\eta =\frac{dW}{Q}=1-\frac{T_C}{T}$

$dW=Q(1-\frac{T_C}{T})$

$dW=c_v(1-\frac{T_C}{T})dT$

integrating From $T_H$ to $T_C$

$\int dW=\int_{T_C}^{T_H}c_v(1-\frac{T_C}{T})dT$

$W=\int_{T_C}^{T_H}33\cdot (1-\frac{300}{T})dT$

$W=c_v\left [ T-T_C\ln T\right ]_{T_H}^{T_C}$

$W=c_v\left [ \left ( T_C-T_H\right )-T_C\left ( \ln \frac{T_C}{T_H}\right )\right ]$

Now heat lost by soup is given by

$Q=c_v(T_C-T_H)$

Fraction of the total heat that is lost by the soup can be turned is given by

$=\frac{W}{Q}$

$=\frac{c_v\left [ \left ( T_C-T_H\right )-T_C\left ( \ln \frac{T_C}{T_H}\right )\right ]}{c_v(T_C-T_H)}$

$=\frac{T_C-T_H-T_C\ln (\frac{T_C}{T_H})}{T_C-T_H}$

$=\frac{300-340-300\ln (\frac{300}{340})}{300-340}$

$=\frac{-40+37.548}{-40}$

$=0.061$

4 0

3 years ago

B) A satellite with mass m orbits the Earth at a radius r. A second satellite also with mass m orbits the

Snezhnost [94]

Answer:

So, given the eqn Fg=G(m1+m2/r^2) where G is the gravitational constant, m is the mass of the satellite and m2 is the mass of the earth and r is the distance from earth to the satellite, the force of earths gravity should be quartered.

Cause (2r)^2 gets turned into (4r^2) where 4r^2 is compared to r^2

Explanation:

6 0

3 years ago