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3 years ago

Which of the following represents the velocity of an object?

1 answer:

8 0

Answer:Velocity can be represented by an arrow, with the length of the arrow representing speed and the way the arrow points representing direction. Objects have the same velocity only if they are moving at the same speed and in the same direction. ... The SI unit for velocity is m/s, plus the direction the object is traveling.

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A battery powers a circuit for a small noisy fan. The fan’s motor gets warm as it turns. What energy transformations are taking

Snowcat [4.5K]

Answer:

The battery is an electric energy source, which is being used to power a motor. The motor turns this electricity into kinetic energy as it spins the fan, creates heat from friction, and creates noise as stated by the problem.

Explanation:

4 0

3 years ago

Read 2 more answers

Light travels from the sun to the earth, a distance of 144,000,000,000 meters, in 480 seconds. Calculate the approximate velocit

mel-nik [20]

Answer:

The answer is C.) 300,000,000 m/s

Explanation:

Light travels at a constant rate of 300,000,000 m/s. this can be determined by dividing the distance (144,000,000,000 meters) by the time (480 seconds). that's pretty fast. in fact, nothing can travel any faster than this. ever.

5 0

3 years ago

A 0.00275 kg air‑inflated balloon is given an excess negative charge q1 =−3.50×10−8 C by rubbing it with a blanket. It is found

kati45 [8]

Answer:

1) $\rm q_2$ is positive.

2) $\rm q_2=4.56\times 10^{-10}\ C.$

Explanation:

The charged rod is held above the balloon and the weight of the balloon acts in downwards direction. To balance the weight of the balloon, the force on the balloon due to the rod must be directed along the upwards direction, which is only possible when the rod exerts an attractive force on the balloon and the electrostatic force on the balloon due to the rod is attractive when the polarities of the charge on the two are different.

Thus, In order for this to occur, the polarity of charge on the rod must be positive, i.e., $\rm q_2$ is positive.

Given:

Mass of the balloon, m = 0.00275 kg.
Charge on the balloon, $\rm q_1 = -3.50\times 10^{-8}\ C.$
Distance between the rod and the balloon, d = 0.0640 m.
Acceleration due to gravity, $\rm g = 9.81\ m/s^2.$

In order to balloon to be float in air, the weight of the balloom must be balanced with the electrostatic force on the balloon due to rod.

Weight of the balloon, $\rm W = mg = 0.00275\times 9.81=2.70\times 10^{-2}\ N.$

The magnitude of the electrostatic force on the balloon due to the rod is given by

$\rm F_e = \dfrac{1}{4\pi \epsilon_o}\dfrac{|q_1||q_2|}{d^2}.$

$\rm \dfrac{1}{4\pi \epsilon_o}$ is the Coulomb's constant.

For the elecric force and the weight to be balanced,

$\rm F_e = W\\\dfrac{1}{4\pi \epsilon_o}\dfrac{|q_1||q_2|}{d^2}=W\\8.99\times 10^9\times \dfrac{3.50\times10^{-8}\times |q_2| }{0.0640^2}=2.70\times 10^{-2}\\|q_2| = \dfrac{2.70\times 10^{-2}\times 0.00640^2}{8.99\times 10^9\times 2.70\times 10^{-7}}=4.56\times 10^{-10}\ C.$

3 0

3 years ago

A 3047.8 kg truck has lost its brakes coming down a mountain. Fortunately, there is a ramp of thick gravel inclined at 9.5 degre

Yuliya22 [10]

Answer:

The work done by the gravel to stop the truck is 520.44 kJ

Explanation:

Step 1: Data given

Mass of the truck = 3047.8 kg

The ramp has an angle of 9.5 °

Velocity of the truck = 20.68 m/s

distance = 26.6 meters

Step 2: Calculate initial kinetic energy

sin 9.5° = 0.165

h = ℓ*sin 9.5° = 26.6*0.165= 4.39 m

Ek = 1/2m*Vo² = 1/2*3047.8*20.68² = 651714.7 Joule = 651.7 kJ = initial kinetic energy

Step 3: Calculate potential energy

Epot = U = m*g*h = 3047.8*9.81*4.39 = 131256.25 Joule = 131.26 kJ

Step 4: What work is done by the truck on the gravel?

Frictional energy Ef = 651.7 kJ - 131.26 kJ = 520.44 kJ

5 0

3 years ago

In case A below, a 1 kg solid sphere is released from rest at point S. It rolls without slipping down the ramp shown, and is lau

mestny [16]

Answer:

the block reaches higher than the sphere

\frac{y_{sphere}} {y_block} = 5/7

Explanation:

We are going to solve this interesting problem

A) in this case a sphere rolls on the ramp, let's find the speed of the center of mass at the exit of the ramp

Let's use the concept of conservation of energy

starting point. At the top of the ramp

Em₀ = U = m g y₁

final point. At the exit of the ramp

Em_f = K + U = ½ m v² + ½ I w² + m g y₂

notice that we include the translational and rotational energy, we assume that the height of the exit ramp is y₂

energy is conserved

Em₀ = Em_f

m g y₁ = ½ m v² + ½ I w² + m g y₂

angular and linear velocity are related

v = w r

the moment of inertia of a sphere is

I = $\frac{2}{5}$ m r²

we substitute

m g (y₁ - y₂) = ½ m v² + ½ ( $\frac{2}{5}$ m r²) ( $\frac{v}{r}$ )²

m g h = ½ m v² (1 + $\frac{2}{5}$ )

where h is the difference in height between the two sides of the ramp

h = y₂ -y₁

mg h = $\frac{7}{5}$ ( $\frac{1}{2}$ m v²)

v = √5/7 √2gh

This is the exit velocity of the vertical movement of the sphere

v_sphere = 0.845 √2gh

B) is the same case, but for a box without friction

starting point

Em₀ = U = mg y₁

final point

Em_f = K + U = ½ m v² + m g y₂

Em₀ = Em_f

mg y₁ = ½ m v² + m g y₂

m g (y₁ -y₂) = ½ m v²

v = √2gh

this is the speed of the box

v_box = √2gh

to know which body reaches higher in the air we can use the kinematic relations

v² = v₀² - 2 g y

at the highest point v = 0

y = vo₀²/ 2g

for the sphere

y_sphere = 5/7 2gh / 2g

y_esfera = 5/7 h

for the block

y_block = 2gh / 2g

y_block = h

therefore the block reaches higher than the sphere

$\frac{y_{sphere}} {y_bolck} = 5/7$

3 0

3 years ago