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2 years ago

15

The magnitude of the vertical velocity vector for an upwardly launched projectile _________. a stays constant b gets smaller and

then larger c decreases throughout the flight d increases throughout the flight

2 answers:

chubhunter [2.5K]2 years ago

7 0

B. Gets smaller and then larger

Masja [62]2 years ago

4 0

Answer:

changes by 9.8 m/s each second.

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The unit light-year is a measure of

stiv31 [10]

A light-year is a unit of distance.

The definition of a light-year is the distance light travels in one year.

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3 years ago

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You push the ball with aforce of 22.8N which induces a -2.3N frictional force. What is the net force while you push?

inna [77]

The net force is 22.8-2.3 or 20.5 N

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3 years ago

A 19 nC charge is moved in a uniform electric field. The electric field does 5.3 μJ of work as the charge moves from point A to

Marizza181 [45]

Answer:

The potential difference between points A and B is 278.95 volts.

The potential difference between points B and C is -642.10 volts.

The potential difference between points A and C is -363.15 volts.

Explanation:

Given :

Charge of the particle, q = 19 nC = 19 x 10⁻⁹ C

Work is done to move a charge from point A to B, W₁ = 5.3 μJ

Work done to move a charge from point B to C, W₂ = -12.2 μJ

Let V₁ be the potential difference between point A and B, V₂ be the potential difference between point B and C and V₃ be the potential difference between point A and C.

The relation between work done and potential difference is:

W = qV

V = W/q ....(1)

Using equation (1), the potential difference between points A and B is:

$V_{1}=\frac{W_{1} }{q}$

Substitute the suitable values in the above equation.

$V_{1} =\frac{5.3\times10^{-6} }{19\times10^{-9} }$

V₁ = 278.95 V

Using equation (1), the potential difference between points B and C is:

$V_{2}=\frac{W_{2} }{q}$

Substitute the suitable values in the above equation.

$V_{2} =\frac{-12.2\times10^{-6} }{19\times10^{-9} }$

V₂ = -642.10 V

The potential difference between points A and C is:

V₃ = V₁ + V₂

V₃ = 278.95 - 642.10

V₃ = -363.15 V

8 0

3 years ago

A bullet with a mass 2.25g is fired up into the air with a velocity of 187.5 m/s. What is the maximum height of the bullet

Ksivusya [100]

Answer:

1793.7m

Explanation:

From the principle of conservation of energy; the kinetic energy substended by the object equals the potential energy sustain by the object when it gets to its maximum position.

Now the kinetic energy; is

K.E = 1/2 × m × v2

Where m is mass

v is velocity

Hence.

K.E = 1/2 × 2.25 × (187.5)^2

Now this should be same with the potential energy which is given as;

P.E = m× g× h

Where m is mass of object

g is acceleration of free fall due to gravity = 9.8m/S2

h is maximum height substain by the object.

Hence P.E = 2.25 × 9.8 × h

From the foregoing analysis of energy conversation it implies;

1/2 × 2.25 × (187.5)^2 =2.25 × 9.8 × h

=> 1/2 × (187.5)^2 = 9.8 × h

=>1/2 × (187.5)^2 / 9.8 = h

=> 1793.69m = h

h= 1793.69m

h =1793.7m to 1 decimal place

3 0

3 years ago

How many centimeters are there in meter? b. 10 c. d 1000 e. 10000 100 2. A centimeter is equal to 1 inch b.½inch C 1/2.54 inch

astra-53 [7]

Answer:

a) There are 100 centimeters in 1 meter.

b) $\texttt{A cm is equal to }\frac{1}{2.54}\texttt{ inch}$

Explanation:

a) We have the conversion

1 m = 100 cm

So there are 100 centimeters in 1 meter.

b) 1 inch = 2.54 cm

$1cm=\frac{1}{2.54}inch$

$\texttt{A cm is equal to }\frac{1}{2.54}\texttt{ inch}$

8 0

3 years ago