Can someone please answer 5-7

Just this last one!!

Pepsi [2]

Answer:

$47 \ \frac{m}{s}$

Explanation:

s = displacement (m)

u = initial velocity $(\frac{m}{s})$

v = final velocity $(\frac{m}{s})$

a = acceleration $(\frac{m}{s^{2} })$

t = time (s)

s = 235

a = -4.7

v = 0

v² = u² + 2as

(0)² = u² + 2(-4.7)(235)

u² - 2209 = 0

u² = 2209

u = 47

4 0

2 years ago

Read 2 more answers

A 20 kg object in a 2 kg object falls toward earth with a acceleration of 9.8m/sWhat is true about the force of gravity on the t

Vitek1552 [10]

Look it up BYEEEEEEEEEEEEEEEEEEEEEE

5 0

3 years ago

The velocity of a 0.25kg model rocket changes from 15m/s [up] to 40m/s [up] in

pochemuha

Since g is constant, the force the escaping gas exerts on the rocket will be 10.4 N

<h3>What is Escape Velocity ?</h3>

This is the minimum velocity required for an object to just escape the gravitational influence of an astronomical body.

Given that the velocity of a 0.25kg model rocket changes from 15m/s [up] to 40m/s [up] in 0.60s. The gravitational field intensity is 9.8N/kg.

To calculate the force the escaping gas exerts of the rocket, let first highlight all the given parameters

Mass (m) of the rocket 0.25 Kg

Initial velocity u = 15 m/s

Final Velocity v = 40 m/s

Time t = 0.6s

Gravitational field intensity g = 9.8N/kg

The force the gas exerts of the rocket = The force on the rocket

The rate change in momentum of the rocket = force applied

F = ma

F = m(v - u)/t

F = 0.25 x (40 - 15)/0.6

F = 0.25 x 41.667

F = 10.42 N

Since g is constant, the force the escaping gas exerts on the rocket is therefore 10.4 N approximately.

Learn more about Escape Velocity here: brainly.com/question/13726115

#SPJ1

7 0

2 years ago

A car traveling on a flat (unbanked) circular track accelerates uniformly from rest with a tangential acceleration of 1.7 m/s^2.

Komok [63]

Answer:

The coefficient of static friction between the car and the track

u=0.572

Explanation:

We don't know the mass of the car or any other information so the acceleration is the reason to solve the friction coefficient

∑ $F=F_{f}=m*a_{t}$

As we know

$F_{f}=u*F_{N}=u*m*g$

Also the center ward direction forces

$F_{fc}=m*a_{c}$

$a_{c}=\frac{v_{t}^2}{r}$

$F_{fc}=m*\frac{v_{t}^2}{r}$

But now vt relation with the tangential acceleration

$v_{t}=2*a_{t}*\frac{\pi }{r}$

replacing

$F_{fc}=m*a_{t}*\frac{2\pi*r}{2r}$

$F_{fc}=m*a_{t}*\pi$

So magnitude of the force can get by

$F_{f}=\sqrt{(m*a_{t}*\pi)^{2}+(m*a_{t})^{2}}$

Get the factor to simplify

$F_{f}=a_{t}*m*\sqrt{(1+\pi^2)}$

$u*m*g=m*a_{t}*(\sqrt{1+\pi^2})$

Solve to u'

$u=\frac{a_{t}}{g}*(\sqrt{1+\pi^2})$

$u=\frac{17\frac{m}{s^2} }{9.8\frac{m}{s^2}}*(\sqrt{1+\pi^2})=0.572$

5 0

3 years ago

a Man is running on straight road with the uniform velocity of 3m per s .calculate acceleration produced by him

Ghella [55]

Answer:

0m/s^2

Explanation:

have a beautiful day ahead

8 0

3 years ago