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3 years ago

What is the impulse of a 1700 kg truck traveling at 22.5 m/s?

1 answer:

3 0

Impulse is just the product of mass and speed!

I = 1700 * 22.5 = 38,250 kg m/s

It does not have any special name for its units :(

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A 375-pound concrete cylinder has a base area of 144 square inches. with the cylinder resting on its base, the pressure exerted

Akimi4 [234]

The pressure exerted by the concrete cylinder is 2.60 pound/in².

We need to know about the pressure to solve this problem. Pressure is a unit that describes how much force is applied to a surface area. It can be determined as

P = F / A

where P is pressure, F is force and A is area.

From the question above, we know that

F = 375 pound

A = 144 in²

By substituting the given parameters, we can calculate the pressure

P = F / A

P = 375 / 144

P = 2.60 pound/in²

Thus, the pressure should be 2.60 pound/in².

Find more on pressure at: brainly.com/question/25965960

#SPJ4

6 0

1 year ago

A uniformly charged ring of radius 10.0 cm has a total charge of 75.0 mC. Find the electric field on the axis of the ring at (a)

wlad13 [49]

Answer:

(a) 6650246.305 N/C

(b) 24150268.34 N/C

(c) 6408227.848 N/C

(d) 665024.6305 N/C

Explanation:

Given:

Radius of the ring (r) = 10.0 cm = 0.10 m [1 cm = 0.01 m]

Total charge of the ring (Q) = 75.0 μC = $75\times 10^{-6}\ \mu C$ [1 μC = 10⁻⁶ C]

Electric field on the axis of the ring of radius 'r' at a distance of 'x' from the center of the ring is given as:

$E_x=\dfrac{kQx}{(x^2+r^2)^\frac{3}{2}}$

Plug in the given values for each point and solve.

(a)

Given:

$Q=75\times 10^{-6}\ \mu C$ , $r=0.01\ m, a=1.00\ cm=0.01\ m,k=9\times 10^{9}\ Nm^2/C^2$

Electric field is given as:

$E_x=\dfrac{(9\times 10^{9})(75\times 10^{-6})(0.01)}{((0.01)^2+(0.1)^2)^\frac{3}{2}}\\\\E_x=\dfrac{6750}{1.015\times 10^{-3}}\\\\E_x=6650246. 305\ N/C$

(b)

Given:

$Q=75\times 10^{-6}\ \mu C$ , $r=0.01\ m, a=5.00\ cm=0.05\ m,k=9\times 10^{9}\ Nm^2/C^2$

Electric field is given as:

$E_x=\dfrac{(9\times 10^{9})(75\times 10^{-6})(0.05)}{((0.05)^2+(0.1)^2)^\frac{3}{2}}\\\\E_x=\dfrac{33750}{1.3975\times 10^{-3}}\\\\E_x=24150268.34\ N/C$

(c)

Given:

$Q=75\times 10^{-6}\ \mu C$ , $r=0.01\ m, a=30.0\ cm=0.30\ m,k=9\times 10^{9}\ Nm^2/C^2$

Electric field is given as:

$E_x=\dfrac{(9\times 10^{9})(75\times 10^{-6})(0.30)}{((0.30)^2+(0.1)^2)^\frac{3}{2}}\\\\E_x=\dfrac{202500}{0.0316}\\\\E_x=6408227.848\ N/C$

(d)

Given:

$Q=75\times 10^{-6}\ \mu C$ , $r=0.01\ m, a=100\ cm=1\ m,k=9\times 10^{9}\ Nm^2/C^2$

Electric field is given as:

$E_x=\dfrac{(9\times 10^{9})(75\times 10^{-6})(1)}{((1)^2+(0.1)^2)^\frac{3}{2}}\\\\E_x=\dfrac{675000}{1.015}\\\\E_x=665024.6305\ N/C$

7 0

3 years ago

Difference between dry cell and simple cell

lakkis [162]

Simple cells have liquid chemicals, making it harder for it to carry. While as dry cells have no liquid chemicals, making it easier to carry.

4 0

3 years ago

What is less dense water or oil steel or water helium or air or oil or water

Leni [432]

Answer:

air because their is nothing contained within the air other than all the solutions that you have listed

Explanation:

4 0

3 years ago

Read 2 more answers

In the above lightwave the property labeled a determines which characteristic of visible light

kap26 [50]

The light must be either very dim or else non-existent.
We can't see the light wave or the label.

7 0

3 years ago