Is my answer correct?

Force (N) = 250<br> Mass (kg) = 221<br> Acceleration (m/s^2) = ?

lyudmila [28]

Answer:

The required acceleration is $a=1.1$ m/s²

Explanation:

Given

Force F = 250 N

Mass m = 221 kg

To determine

Acceleration a = ?

We know that acceleration is produced when a force is applied to a body.

The acceleration can be determined using the formula

$F = ma$

where

F is the force

m is the mass

a is the acceleration

now substituting F = 250 , and m = 221 in the formula

$F = ma$

$250 = 221 (a)$

switch the equation

$221(a) = 250$

Divide both sides by 221

$\frac{221a}{221}=\frac{250}{221}$

simplify

$a=\frac{250}{221}$

$a=1.1$ m/s²

Therefore, the required acceleration is $a=1.1$ m/s²

8 0

3 years ago

C. A rocket moves from the velocity

Pavlova-9 [17]

Answer:

20s

Explanation:

t=d/(V2-V1)

t= 80000m/(6000-2000)

t= 20s

6 0

2 years ago

The charge density of radius 12.0 cm isgiven by with a = 1.40 μC/m 3 and r ismeasured radially outward from theorigin. What is t

Zigmanuir [339]

Answer:

The electric potential of the uniformly charge disk is 1392.1 V

Explanation:

Electric potential, for a uniformly charged disk at a distance A, is given as;

$V = \frac{\sigma}{2 \epsilon} [\sqrt{A^2 +R^2} -A]$

Where;

σ is the charge density = 1.40 μC/m³

ε is the permittivity of free space = 8.85 x 10⁻¹²

A is the distance above the disk = 40 cm = 0.4 m

R is the radius of the disk = 0.12 m

Substitute in these values into the equation above, we will have

$V = \frac{1.4 X 10^{-6}}{2X8,85X10^{-12}}[\sqrt{0.4^2 +0.12^2}-0.4] \\\\V = (79096.05)(0.0176) = 1392.1 V$

Therefore, the electric potential of the uniformly charge disk is 1392.1 V

3 0

3 years ago

5. What is the difference between an object's SPEED and its VELOCITY?

Svetach [21]

Speed is the time rate at which an object is moving along a path, while velocity is the rate and direction of an object's movement.

3 0

3 years ago

During your summer internship for an aerospace company, you are asked to design a small research rocket. The rocket is to be lau

Thepotemich [5.8K]

Answer:

$T=6.75s$

Explanation:

We must separate the motion into two parts, the first when the rocket's engines is on and the second when the rocket's engines is off. So, we need to know the height ( $h_1$ ) that the rocket reaches while its engine is on and we need to know the distance ( $h_2$ ) that it travels while its engine is off.

For solving this we use the kinematic equations:

In the first part we have:

$h_1=v_0T+\frac{1}{2}aT^2\\h_1=0*T+\frac{1}{2}(16\frac{m}{s^2})T^2\\h_1=8\frac{m}{s^2}T^2\\$

and the final speed is:

$v_f=v_0+aT\\v_f=0+16\frac{m}{s^2}T\\v_f=16\frac{m}{s^2}T$

In the second part, the final speed of the first part it will be the initial speed, and the final speed is zero, since gravity slows it down the rocket.

So, we have:

$v_f^2=v_0^2+2gh_2\\2gh_2=v_f^2-v_0^2\\h_2=\frac{v_f^2-v_0^2}{2g}\\h_2=\frac{0^2-(16\frac{m}{s^2}T)^2}{2(-9.8\frac{m}{s^2})}\\h_2=\frac{-256\frac{m^2}{s^4}T^2}{-19.6\frac{m}{s^2}}\\h_2=13.06\frac{m}{s^2}T^2$

The sum of these heights will give us the total height, which is known:

$h=h_1+h_2\\960m=8\frac{m}{s^2}T^2+13.06\frac{m}{s^2}T^2\\960m=21.06\frac{m}{s^2}T^2\\T^2=\frac{960m}{21.06\frac{m}{s^2}}\\T^2=45.58s^2\\T=\sqrt{45.58s^2}\\T=6.75s$

This is the time that its needed in order for the rocket to reach the required altitude.

5 0

3 years ago