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storchak [24]
3 years ago
11

Could I get some help with this

Physics
1 answer:
nika2105 [10]3 years ago
4 0

please give me brainlest!!

the answer is A.

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A student made the claim that a 4 gram paintball fired from a paintball gun at 90 m/s could have about the same kinetic energy a
vovikov84 [41]

This question involves the concept of kinetic energy.

The student's claim is "right".

<h3>Kinetic Energy</h3>

The energy possessed by a body, by the virtue of its motion is called kinetic energy. Mathematically it is given by the following formula:

K.E =\frac{1}{2}mv^2

where,

  • K.E = Kinetic energy
  • m = mass
  • v = velocity

Therefore,

For the paintball:

K.E = \frac{1}{2}(4\ g)(90\ m/s)^2

K.E = 16200 J

For the pellet:

K.E = \frac{1}{2}(1\ g)(180\ m/s)^2

K.E = 16200 J

Hence, both paintball and pellet will have same kinetic energy. The student is right.

Learn more about kinetic energy here:

brainly.com/question/12669551

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7 0
2 years ago
A grandfather clock works by swinging a pendulum back and forth with a
allochka39001 [22]

Answer:

For example, a wave with a time period of 2 seconds has a frequency of

1 ÷ 2 = 0.5 Hz.

Explanation:

8 0
3 years ago
Use the worked example above to help you solve this problem. An Alaskan rescue plane drops a package of emergency rations to str
Vlad [161]

Answer:

(a) The package lands 682 meters horizontally ahead from the point the package was dropped from the plane

(b) The horizontal component = 39.0 m/s

The vertical component = 171.55 m/s

(c) The angle of impact is 77.19°

Explanation:

The parameters given are;

Velocity of the plane, vₓ = 39.0 m/s

Height of the plane above the ground, h = 1.50 × 10² m = 1,500 m

(a) The time, t, before the package hits the ground when dropped from the plane is given by the relation;

h = u·t + 1/2×g×t²

Where:

g = Acceleration due to gravity = 9.81 m/s²

u = Initial vertical velocity = 0 m/s

Hence;

1500 = 0×t + 1/2 × 9.81 × t² = 4.905·t²

∴ t = √(1500/4.905) = 17.49 s

The horizontal distance the package travels before landing = 17.49 × 39 ≈ 682 m

The package lands 682 meters horizontally ahead from the point the package was dropped from the plane

(b) The vertical velocity, v_y, of the package just before landing is given by the relation;

v_y² = u² + 2·g·h

u = 0 m/s

∴ v_y² = 0 + 2×9.81×1500 = 29430 m²/s²

v_y = √29430  = 171.55 m/s

Hence the horizontal component = 39.0 m/s

The vertical component = 171.55 m/s

(c) The angle of impact, θ, is given as follows;

tan \theta = \dfrac{v_y}{v_x}  = \dfrac{171.55}{39.0 } = 4.4

∴ θ = tan⁻¹(4.4) = 77.19°.

6 0
4 years ago
Which of the following is the correct SI unit to measure of rope?
sergij07 [2.7K]

Hello!

The answer should be the third option.

Explanation: Because the meters is the correct SI unit to measure of rope. And thank you for posting at here on Brainly. And have a great day! -Charlie

7 0
3 years ago
A 65-kg skier grips a moving rope that is powered by an engine and is pulled at a constant speed to the top of a 230 hill. The s
Sveta_85 [38]

Answer:

The required power by the engine is 33.0 hp

Explanation:

Solution

Newton's second law says that, the net force Fnet on an object of mass m will accelerates the object

Where

Fnet = ma

a = acceleration

θ = angle of incline,

m = mass of the 30 skiers,

f = frictional force

N = normal force

mg sinθ, mg cos θ are components of weight skier

F = the force applied by engine

Now,

The skier mass  is 65 kg

We calculate the mass of the 30 skier

m = 30 (65kg) = 1950 kg

Calculate the net force acting on the skiers along the x-axis

Fnet, x=ma

Now,

F-mg sin θ - f = 0

F= mg sin θ + f -----(1)

The kinetic frictional force is denoted by

f = μk N ------(2)

μk = The coefficient of the kinetic friction

We now, calculate the net force acting on the skiers along y axis

Fnet, y = ma

N- mg cos θ = 0

so,

N = mg cos θ

This value is  substituted in equation (2)

f = μk mg cos θ

we substitute the value for equation (1)

F = mg sin θ + μk mg cos θ

mg =  sinθ + μk cos θ)-----(3)

The next step is to calculate the work done by the engine in pulling the skiers, the incline top by applying the equation 3

W = Fx

= mg ( sinθ + μk cos θ)x

x = the displacement

we now substitute 1950 kg for m, 23° for θ, 0.10 for μk and 320m for x

so,

W = mg ( sinθ + μk cos θ)x

= (1950 kg) (9.81 m/s²) (sin 23° + (0.10) cos 23°) (320 m)

= 2.99 * 10 ^6 J

Then,

The time from minute to s is converted

t =(2.0min) ( 60sec/1.0min) = 120 sec

Now we calculate the power needed by the engine to pull the skiers at the incline top

Thus,

P = W/t

we substitute  2.955 * 10 ^6 J for W and  120 s for t

we have,

P = 2.955 * 10 ^ 6 J/ 120 s

= ( 2.4625 * 10 ^ 4 W) (1.0 hp/746 W)

= 33.0 hp

In conclusion the required power by the engine is 33.0 hp

3 0
3 years ago
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