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3 years ago

energy transfered electrically is measured in _____. what one word completes the sentence?give the fullname not symbol.

Physics

1 answer:

Dmitry [639]3 years ago

4 0

Answer:

joules

Explanation:

The electric energy transferred to a resistor in a time period is equal to the electric power multiplied by time, E=Pt, and can also be calculated using E=I2Rt. Electric companies measure their energy sales in a large number of joules called a kilowatt hour (kWh) which is equivalent to 3.6×106 J.

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A cardinal (Richmondena cardinalis) of mass 4.00×10−2 kg and a baseball of mass 0.146 kg have the same kinetic energy. What is t

sp2606 [1]

Answer:

0.5232

Explanation:

the cardinal and the baseball has the same kinetic energy

$\frac{1}{2} m_{c} v_{c} ^{2} = \frac{1}{2} m_{b} v_{b} ^{2}$

$m_{c}v_{c}^{2} = m_{b}v_{b}^{2}$

$[4.0X 10^{-2} ]v_{c} ^{2} = [0.146]v_{b} ^{2}$

$\frac{v_{c} }{v_{b} } = \sqrt{\frac{0.146}{4.0 X 10^{-2} } }$

$\frac{v_{c} }{v_{b} } = 1.910$

Ratio of momentum

$\frac{p_{c}}{p_{b}} = \frac{m_{c}v_{c}}{m_{b}v_{b}} = \frac{ (4.0 X 10^{-2} )(1.910v_{b} )}{0.146v_{b} }$

$\frac{p_{c}}{p_{b}} = \frac{4.0 X 10 ^{-2} X 1.910 }{0.146} = 0.5232$

6 0

4 years ago

If a CFOC was launched and travels 65 meters and is in the air for 3 seconds, what is the launch velocity and angle?

labwork [276]

Answer:

Lauch velocity (u) = 26.15 m/s

Lauch Angle (θ) = 35°

Explanation:

From the question given above, the following data were obtained:

Range (R) = 65 m

Time of flight (T) = 3 s

Acceleration due to gravity (g) = 10 m/s²

Lauch velocity (u) =?

Lauch Angle (θ) =?

R = u²Sin2θ /g

65 = u² × Sin2θ /10

Recall:

Sin2θ = 2SinθCosθ

65 = u² × 2SinθCosθ / 10

65 = u² × SinθCosθ / 5

Cross multiply

65 × 5 = u² × SinθCosθ

325 = u² × SinθCosθ .....(1)

T = 2uSinθ / g

3 = 2uSinθ / 10

3 = uSinθ / 5

Cross multiply

3 × 5 = uSinθ

15 = u × Sinθ

Divide both side by Sinθ

u = 15 / Sinθ....... (2)

Substitute the value of u in equation (2) into equation (1)

325 = u² × SinθCosθ

u = 15 / Sinθ

325 = (15 / Sinθ)² × SinθCosθ

325 = 225 / Sin²θ × SinθCosθ

325 = 225 × SinθCosθ / Sin²θ

325 = 225 × Cosθ / Sinθ

Cross multiply

325 × Sineθ = 225 × Cosθ

Divide both side by Cosθ

325 × Sineθ / Cosθ = 225

Divide both side by 325

Sineθ / Cosθ = 225 / 325

Sineθ / Cosθ = 0.6923

Recall:

Sineθ / Cosθ = Tanθ

Tanθ = 0.6923

Take the inverse of Tan

θ = Tan¯¹ 0.6923

θ = 35°

Substitute the value of θ into equation (2) to obtain the value of u.

u = 15 / Sinθ

θ = 35°

u = 15 / Sin 35

u = 15 / 0.5736

u = 26.15 m/s

Summary:

Lauch velocity (u) = 26.15 m/s

Lauch Angle (θ) = 35°

8 0

3 years ago

Name the types of motion represented by oA, AB and BC .

aivan3 [116]

Answer:

OA-Uniformly accelerated motion.
AB-constant speed,hence uniform motion.
BC-motion with uniform (-)ye acceleration.

8 0

3 years ago

A CD has a mass of 17 g and a radius of 6.0 cm. When inserted into a player, the CD starts from rest and accelerates to an angul

Stells [14]

Answer:

τ =9.41 * 10⁻⁴ N*m

Explanation:

Kinematics of the CD

The CD rotates with constant angular acceleration and its angular acceleration is calculated as follows:

$\alpha = \frac{\omega_{f}- \omega_{i}}{t}$ Formula (1)

Where:

α : angular acceleration. (rad/s²)

ωf: final angular velocity (rad/s)

ωi : initial angular velocity (rad/s)

t = time interval (s)

Data

ωf= 20 rad/s

ωi =0

t = 0.65 s.

Calculating of the angular acceleration of the CD

We replace data in the formula (1)

$\alpha = \frac{20 -0}{0.65}$

α = 30.77 rad/s²

Newton's second law in rotation:

F = ma has the equivalent for rotation:

τ = I * α Formula (2)

where:

τ : It is the net torque applied to the body. (N*m)

I : it is the moment of inertia of the body with respect to the axis of rotation (kg*m²)

α : It is angular acceleration. (rad/s²)

Calculating of the moment of inertia of the CD

The moment of inertia of a disk with respect to an axis perpendicular to the plane and passing through its center is calculated by the following formula:

I = (1/2) M*R² Formula (3)

Data

M= 17 g = 17/1000 kg = 0.017 kg : CD mass

R= 6.0 cm= 6/100 m = 0.06 m : CD radius

We replace data in the formula (3) :

I = (1/2) ( 0.017 kg)*(0.06 m)² = 3.06 * 10⁻⁵ kg*m²

Calculating of the net torque acting on the CD

Data

α = 30.77 rad/s²

I = 3.06 * 10⁻⁵ kg*m²

We replace data in the formula (2) :

τ = I * α

τ =( 3.06 * 10⁻⁵ kg*m²) * (30.77 rad/s²)

τ =9.41 * 10⁻⁴ N*m

3 0

3 years ago

Important parts of stydi g physics

sladkih [1.3K]

Answer:

put your question in proper way' i hope you understand.

Explanation:

7 0

2 years ago