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Anettt [7]
3 years ago
12

A ball is thrown upward from the edge of a cliff with an initial velocity of 6 m/s. (a) How fast is it moving 0.5 s later? In wh

at direction? (b) How fast is it moving 2 s later? In what direction?​
Physics
1 answer:
almond37 [142]3 years ago
7 0

Answer:

Explanation:

Kinematic equation

v = u + at

If UP is assumed to be the positive direction and we let gravity be 10 m/s² which will be in the downward direction so will be negative.

a) v = 6  + (-10)(0.5) = 1 m/s    the result is positive, so upward

b) v = 6  + (-10)(2) = -14 m/s    the result is negative, so downward

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A 60-W, 120-V light bulb and a 200-W, 120-V light bulb are connected in series across a 240-V line. Assume that the resistance o
gulaghasi [49]

A. 0.77 A

Using the relationship:

P=\frac{V^2}{R}

where P is the power, V is the voltage, and R the resistance, we can find the resistance of each bulb.

For the first light bulb, P = 60 W and V = 120 V, so the resistance is

R_1=\frac{V^2}{P}=\frac{(120 V)^2}{60 W}=240 \Omega

For the second light bulb, P = 200 W and V = 120 V, so the resistance is

R_1=\frac{V^2}{P}=\frac{(120 V)^2}{200 W}=72 \Omega

The two light bulbs are connected in series, so their equivalent resistance is

R=R_1 + R_2 = 240 \Omega + 72 \Omega =312 \Omega

The two light bulbs are connected to a voltage of

V  = 240 V

So we can find the current through the two bulbs by using Ohm's law:

I=\frac{V}{R}=\frac{240 V}{312 \Omega}=0.77 A

B. 142.3 W

The power dissipated in the first bulb is given by:

P_1=I^2 R_1

where

I = 0.77 A is the current

R_1 = 240 \Omega is the resistance of the bulb

Substituting numbers, we get

P_1 = (0.77 A)^2 (240 \Omega)=142.3 W

C. 42.7 W

The power dissipated in the second bulb is given by:

P_2=I^2 R_2

where

I = 0.77 A is the current

R_2 = 72 \Omega is the resistance of the bulb

Substituting numbers, we get

P_2 = (0.77 A)^2 (72 \Omega)=42.7 W

D. The 60-W bulb burns out very quickly

The power dissipated by the resistance of each light bulb is equal to:

P=\frac{E}{t}

where

E is the amount of energy dissipated

t is the time interval

From part B and C we see that the 60 W bulb dissipates more power (142.3 W) than the 200-W bulb (42.7 W). This means that the first bulb dissipates energy faster than the second bulb, so it also burns out faster.

7 0
3 years ago
What happens when a star blows up and it is next another star will it blow up too?
FinnZ [79.3K]

Answer:

Sometimes

Explanation:

Sometimes meaning in occasions. If it is a "vampire star" not likely because they also cause they also make novas happen by sucking the gas from another star.

4 0
4 years ago
Read 2 more answers
Paco was driving his scooter west with an initial velocity of 4 m/s. He accelerates at 0.5 m/s2 for 30 seconds.
KonstantinChe [14]

Answer:

V = 19m/s

Explanation:

Given the following data;

Initial velocity, U = 4m/s

Acceleration, a = 0.5m/s²

Time, t = 30 seconds

To find the final velocity, we would use the first equation of motion;

V = U + at

Where;

V is the final velocity.

U is the initial velocity.

a is the acceleration.

t is the time measured in seconds.

V = 4 + 0.5*30

V = 4 + 15

V = 19m/s

Therefore, his final velocity is 19 meters per seconds.

6 0
3 years ago
Read 2 more answers
A spring on a horizontal surface can be stretched and held 0.5 m from its equilibrium position with a force of 60 N. a. How much
Lostsunrise [7]

Answer:

a)1815Joules b) 185Joules

Explanation:

Hooke's law states that the extension of a material is directly proportional to the applied force provided that the elastic limit is not exceeded. Mathematically;

F = ke where;

F is the applied force

k is the elastic constant

e is the extension of the material

From the formula, k = F/e

F1/e1 = F2/e2

If a force of 60N causes an extension of 0.5m of the string from its equilibrium position, the elastic constant of the spring will be ;

k = 60/0.5

k = 120N/m

a) To get the work done in stretching the spring 5.5m from its position,

Work done by the spring = 1/2ke²

Given k = 120N/m, e = 5.5m

Work done = 1/2×120×5.5²

Work done = 60× 5.5²

Work done = 1815Joules

b) work done in compressing the spring 1.5m from its equilibrium position will be gotten using the same formula;

Work done = 1/2ke²

Work done =1/2× 120×1.5²

Works done = 60×1.5²

Work done = 135Joules

8 0
3 years ago
Read 2 more answers
If you know how much radioactive material the organism had to begin with, explain how you could use half-life to determine its a
TiliK225 [7]

Answer: We can calculate it with the radioactive half life equation

Explanation:

If we already know the initial amount of radioactive material and its half life, we can leave that material for a specific known time and then measure how much of the material is left (since it follows the radioactive deacay) and use the results in the following formula:

A=A_{o}.2^{\frac{-t}{h}}  

Where:

A is the final amount of the material

A_{o} is the initial amount of the material

t is the time elapsed

h is the half life of the radioactive compound

5 0
3 years ago
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