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4 years ago

If your speed slows down you will have what acceleration?

Physics

2 answers:

vitfil [10]4 years ago

8 0

When speed slows down you have kinetic energy.

Mrac [35]4 years ago

4 0

If you slow down you have deceleration

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How fast is a 950 kg car traveling if it has 3,800 Joules of energy

inysia [295]

Answer:

The kinetic energy K of a moving object is: K= 1/2mv to the second power

where m is the mass and v the velocity of the object.

Using this formula, we can calculate v for this problem

Explanation:

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3 years ago

A 3.0\, \text {kg}3.0kg3, point, 0, start text, k, g, end text box is at rest on a table. The static friction coefficient \mu_sμ

Kipish [7]

Answer:

its 0 m/s^2 on khan

Explanation:

usually how i think of it if its below 15 N then its 0

3 0

3 years ago

Read 2 more answers

A metal sphere of radius 11 cm has a net charge of 2.8 × 10–8 C. (a) What is the electric field at the sphere's surface? (b) If

harina [27]

Answer:

Therefore,

a) $E=20.82\ kN/C$

b) $V= 2290\ Volt$

c) distance from the sphere's surface = 1.8 cm

Explanation:

Given:

Radius, r = 11 cm = 0.11 m

Charge, $Q = 2.8\times 10^{-8}\ C$

To Find:

a) electric field at the sphere's surface = ?

b) If V = 0 at infinity, what is the electric potential at the sphere's surface = ?

c) At what distance from the sphere's surface has the electric potential decreased by 320 V = ?

Solution:

Electric field at the sphere's surface is given as,

$E=\dfrac{k\times Q}{r^{2}}$

Where,

E = Electric Field,

$k = Coulomb's\ constant = 9\times 10^{9}$

Q = Charge

r = Radius

Substituting the values we get

$E=\dfrac{9\times 10^{9}\times 2.8\times 10^{-8}}{(0.11)^{2}}=20.82\ kN/C$

Now, Electric Potential at point surface is given as,

$V=\dfrac{k\times Q}{r}$

Substituting the values we get

$V=\dfrac{9\times 10^{9}\times 2.8\times 10^{-8}}{0.11}=2.29\ kV$

For distance from the sphere's surface has the electric potential decreased by 320 V,

So V becomes,

V = 2290 - 320 = 1970 Volt, then r =?

$V=\dfrac{k\times Q}{r}$

Substituting the values we get

$r=\dfrac{k\times Q}{V}\\\\r=\dfrac{9\times 10^{9}\times 2.8\times 10^{-8}}{1970}=0.128\ m$

Therefore the distance from the sphere's surface,

$d = 12.8 - 11 = 1.8\ cm$

Therefore,

a) $E=20.82\ kN/C$

b) $V= 2290\ Volt$

c) distance from the sphere's surface = 1.8 cm

7 0

4 years ago

if an object has a velocity of 165 miles per hour, how many hours will it take to travel 5,782 miles, to the nearest hour

lisov135 [29]

34 to the nearest hour

7 0

3 years ago

Read 2 more answers

You calibrated a different pipet and determined its volume to be 25.40 mL. You then use it to determine the density of an unknow

just olya [345]

Answer:

1148.37401575 kg/m³

Explanation:

Volume of liquid in pipet

25.4 mL

Mass of liquid

$77.8760-48.7073=29.1687\ g$

Density is given by

$\rho=\dfrac{m}{V}\\\Rightarrow \rho=\dfrac{29.1687}{25.4}\\\Rightarrow \rho=1.14837401575\ g/ml=1.14837401575\times 10^3=1148.37401575\ kg/m^3$

The density of the unknown liquid is 1148.37401575 kg/m³

4 0

3 years ago