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3 years ago

Can someone pls help, thank you in advance!

Physics

1 answer:

zzz [600]3 years ago

7 0

Answer:

an object sliding down hill

Explanation:

On a slope, the force applied is due to gravity. Its direction is straight down. If the object is sliding down the hill, its displacement is at an angle to the applied force. The angle of displacement will depend on the steepness of the hill.

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Some curious students hold a rolling race by rolling four items down a steep hill. The four items are a solid homogeneous sphere

Papessa [141]

Answer

The answer and procedures of the exercise are attached in the following archives.

Step-by-step explanation:

You will find the procedures, formulas or necessary explanations in the archive attached below. If you have any question ask and I will aclare your doubts kindly.

8 0

3 years ago

What ia the law of convesation of energy

Free_Kalibri [48]

In physics and chemistry, the law of conservation of energy states that the total energy of an isolated system remains constant; it is said to be conserved over time. This law means that energy can neither be created nor destroyed; rather, it can only be transformed or transferred from one form to another.

8 0

3 years ago

A current of 9 A flows through an electric device with a resistance of 43 Ω. What must be the applied voltage in this particular

lubasha [3.4K]

Voltage, V = IR

Where I is current in Ampere, R is Resistance in Ohms.

V = 9A * 43 Ω

V = 387 V

8 0

3 years ago

Please help Need good grade

ivann1987 [24]

Answer:

The other angle is 120°.

Explanation:

Given that,

Angle = 60

Speed = 5.0

We need to calculate the range

Using formula of range

$R=\dfrac{v^2\sin(2\theta)}{g}$ ...(I)

The range for the other angle is

$R=\dfrac{v^2\sin(2(\alpha-\theta))}{g}$ ....(II)

Here, distance and speed are same

On comparing both range

$\dfrac{v^2\sin(2\theta)}{g}=\dfrac{v^2\sin(2(\alpha-\theta))}{g}$

$\sin(2\theta)=\sin(2\times(\alpha-\theta))$

$\sin120=\sin2(\alpha-60)$

$120=2\alpha-120$

$\alpha=\dfrac{120+120}{2}$

$\alpha=120^{\circ}$

Hence, The other angle is 120°

6 0

3 years ago

A(n) 55.5 g ball is dropped from a height of 53.6 cm above a spring of negligible mass. The ball compresses the spring to a maxi

Serggg [28]

Answer:

The spring force constant is $k=243\ \frac{N}{m}$ .

Explanation:

We are told the mass of the ball is $m=0.0555\ kg$ , the height above the spring where the ball is dropped is $h=0.536\ m$ , the length the ball compresses the spring is $d=0.04897\ m$ and the acceleration of gravity is $9.8\ \frac{m}{s^{2}}$ .

We will consider the initial moment to be when the ball is dropped and the final moment to be when the ball stops, compressing the spring. We supose that there is no friction so the initial mechanical energy $E_{mi}$ is equal to the final mechanical energy $E_{mf}$ :

$E_{mf}=E_{mi}$

Initially there is only gravitational potential energy because the force of the spring isn't present and the speed is zero. In the final moment there is only elastic potential energy because the height is zero and the ball has stopped. So we have that:

$\frac{1}{2}kd^{2}=mgh$

If we manipulate the equation we have that:

$k=\frac{2mgh}{d^{2} }$

$k=\frac{2\ 0.0555\ kg\ 9.8\frac{m}{s^{2}}\ 0.536\ m}{(0.04897)^{2}m^{2}}$

$k=\frac{0.58306\ \frac{kgm^{2}}{s^{2}}}{2.398x10^{-3}m^{2}}$

$k=243\ \frac{N}{m}$

5 0

3 years ago