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choli [55]
3 years ago
9

Suppose that a baseball is tossed up into the air at an initial velocity 1818​​\text{ m/s} m/s​. The height of the baseball at t

ime tt​ in seconds is given by h\mathopen{}\mathclose{\left(t\right)}=18 t-4.9 t^{2}h(t)=18t−4.9t 2 ​​​​​ (in meters). a) What is the average velocity for \left[1, 1.5\right][1,1.5]​?
Physics
1 answer:
olga2289 [7]3 years ago
5 0

Answer:

13.1 m/s

Explanation:

Given that a baseball is tossed up into the air at an initial velocity 18 m/s​. The height of the baseball at time t​ in seconds is given by h(t) = 18t−4.9t 2 ​​​​​ (in meters).

a) What is the average velocity for [1,1.5]​?

To calculate the velocity travelled by the ball, differentiate the function.

dh/dt = 18 - 9.8t

Substitute t for 1 in the above Differential function

dh/dt = 18 - 9.8 (1)

But dh/dt = velocity

V = 18 - 9.8

V = 8.2 m/s

Average velocity = ( U + V ) / 2

Average velocity = (18 + 8.2)/2

Average velocity = 26.2/2

Average velocity = 13.1 m/s

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djverab [1.8K]

The answer is A. Determining the order of rock layers

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4 years ago
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What distance was an object moved by a force of 40 N if the work was 600 joules
Westkost [7]

If a force of 40 N does 600 joules of work, then it must have been applied continuously through a distance of  (600/40) = 15 meters .

We can't tell how far the object ultimately moved, because we don't know
what other forces may have been acting on it.  If there were NO other forces,
then it just kept going, even after the 40-N force stopped, with the same
600 joules of kinetic energy that it got from the work that the force did.


6 0
3 years ago
Read 2 more answers
Suppose that a car starts from rest at t = 0. The car moves with an acceleration of 1.5 m/s2. How far will the car travel in 3.0
rodikova [14]

Answer:

6.75m

Explanation:

To calculate the distance in this question, we can use the formula:

S = ut + 1/2at^2

Where; S = distance

u = initial velocity = 0m/s

t = 3s

a = 1.5m/s^2

Hence:

S = (0 × 3) + 1/2 (1.5 × 3 × 3)

S = 0 + 1/2 (13.5)

S = 13.5/2

S = 6.75

Therefore, the car will travel 6.75m in 3seconds.

3 0
4 years ago
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
A lightbulb has a resistance of 195 Ω and carries a current of 0.62 A.
ki77a [65]

Answer:

75

Explanation:

Power is current times voltage:

P = IV

Voltage is current times resistance:

V = IR

Therefore:

P = I²R

Given I = 0.62 A and R = 195 Ω:

P = (0.62 A)² (195 Ω)

P ≈ 75 W

8 0
3 years ago
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