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Arada [10]
3 years ago
6

How does the speed of an object relate to the energy of the object

Physics
1 answer:
Ann [662]3 years ago
7 0

Answer:

Increasing the speed of an object decreases its motion energy. Increasing the speed of an object increases its motion energy. Increasing the speed of an object does not affect its motion energy. Whether or not its motion energy is affected depends on how much its speed was increased.

Explanation:

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Which situation describes the highest rate of power?
Darya [45]

Answer : The correct option is, (D) A machine does 400 joules of work in 5 seconds.

Explanation :

Power : It is defined a the rate of doing work per unit time.

Formula used :

P=\frac{w}{t}

where,

P = power

w = work done

t = time

Now we have to determine the rate of power for the following options.

(A) A machine does 200 joules of work in 10 seconds.

P=\frac{200}{10}=20W

(B) A machine does 400 joules of work in 10 seconds.

 P=\frac{400}{10}=40W

(C) A machine does 200 joules of work in 5 seconds.

 P=\frac{200}{5}=40W

(D) A machine does 400 joules of work in 5 seconds.

P=\frac{400}{5}=80W

From this we conclude that, a machine does 400 joules of work in 5 seconds has the highest rate of power.

Hence, the correct option is, (D)

8 0
3 years ago
After your school's team wins the regional championship, students go to the dorm roof and start setting off fireworks rockets. T
PIT_PIT [208]

Answer:

160 m

Explanation:

The intensity, I, of the sound is inversely proportional to the square of the distance, r, from the source.

I\propto \dfrac{1}{r^2}

Hence,

I_1r_1^2 = I_2r_2^2

r_2 = r_1\sqrt{\dfrac{I_1}{I_2}}

From the question, I_2 is half of I_1

r_2 = r_1\sqrt{\dfrac{I_1}{0.5I_1}}

r_2 = r_1\sqrt{2}

r_2 = 113\text{ m}\sqrt{2} = 160 \text{ m}

3 0
3 years ago
Read 2 more answers
A coin 15.0 mm in diameter is placed 15.0 cm from a spherical mirror. The coin's image is 5.0 mm in diameter and is erect. Is th
s344n2d4d5 [400]

Answer:

15 cm

Explanation:

h_{o} = Diameter of the coin = 15 mm

h_{i} = Diameter of the image of coin =  5 mm

d_{o} = distance of the coin from mirror = 15 cm

d_{i} = distance of the image of coin from mirror = ?

Using the equation

\frac{d_{i}}{d_{o}} = \frac{- h_{i}}{h_{o}}

\frac{d_{i}}{15} = \frac{- (5)}{15}

d_{i} = - 5 cm

R = radius of curvature

Using the mirror equation

\frac{1}{d_{o}} + \frac{1}{d_{i}} = \frac{2}{R}

\frac{1}{15} + \frac{1}{- 5} = \frac{2}{R}

R = - 15 cm

6 0
3 years ago
represents the space-time, speed-time and acceleration-time graphs for a ball pulled upwards with a speed of 10 m / s from 1 met
alexandr1967 [171]
Gyhugugygubuhuhuhubu
5 0
3 years ago
Light is shone on a diffraction grating
Pani-rosa [81]

Answer:

    λ = 482.05 nm

Explanation:

The diffraction phenomenon and the diffraction grating is described by the expression

         d sin θ = m λ

where d is the distance between two consecutive slits, λ the wavelength and m an integer representing the order of diffraction

in this case they indicate the distance between slits, the angle and the order of diffraction

         λ = \frac{d sin \theta }{m}d sin θ / m

let's calculate

         λ = 1.00 10⁻⁶ sin 74.6 / 2

         λ = 4.82048 10⁻⁷ m

Let's reduce to nm

         λ = 4.82048 10⁻⁷ m (10⁹ nm / 1 m)

         λ = 482.05 nm

3 0
3 years ago
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