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2 years ago

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A ball of mass 0.35 kg hangs straight down on a string of length 13 cm. It is then swung upward, keeping the string taut, until

the string makes an angle of 70° with respect to the vertical. Find the change in the gravitational potential energy of the Earth-ball system.

1 answer:

Anni [7]2 years ago

6 0

Answer:

PE=0.29J

Explanation:

According to the description, there is a angle and in point swung upward of 70°

So,

$Y=13*10^{-2}*cos(70) \\Y=0.0444m$

Appling the equation of Potential Energy we have,

$PE=mgh\\PE=(0.35)(9.8)(0.13-0.0444)\\PE=0.29J$

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What causes an electrical current in a wire? Question 1 options: Electric fields cause atoms to move in a wire Electrons build u

Mademuasel [1]

The cause of electric current in a wire is that there is electron flow due to electrical attraction and repulsion.

What is Electric current?

The rate of flow of electric charge through a point is known as electric current. The electric current is nothing but the flow of electrons in general.

Consider a space, when electrons of charge (e) is flowing through the space in a given time interval t. Then,

I = e/t

here,

I is the electric current.

Moreover, there is a flow of electrons in a wire due to a potential difference existing at the circuit. Due to that, electrons always flow from high electric potential to low electric potential.

Thus, we can conclude that the cause of electric current in a wire is that there is electron flow due to electrical attraction and repulsion.

Learn more about the electric current here:

brainly.com/question/7643273

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2 years ago

A mass m = 3.9 kg hangs from a massless string wrapped around a uniform cylinder with mass Mp = 10.53 and radius R= 0.96 m. The

luda_lava [24]

Answer:

the rotational inertia of the cylinder = 4.85 kgm²

the mass moved 7.942 m/s

Explanation:

Formula for calculating Inertia can be expressed as:

$I =\frac{1}{2}mR^2$

For calculating the rotational inertia of the cylinder ; we have;

$I = \frac{1}{2}m_pR^2$

$I = \frac{1}{2}*10.53*(0.96)^2$

$I=5.265*(0.96)^2$

$I=4.852224$

I ≅ 4.85 kgm²

mg - T ma and RT = I ∝

T = $\frac{Ia}{R^2}$

$a = \frac{g}{1+\frac{I}{mR^2}}$

$a = \frac{9.8}{1+\frac{4.85}{3.9*(0.96)^2}}$

a = 4.1713 m/s²

Using the equation of motion

$v^2 = u^2+2as \\ \\ v^2 = 2as \\ \\ v = \sqrt{2*a*s} \\ \\ v= \sqrt{2*4.1713*7.56} \\ \\ v = 7.942 \ m/s$

3 0

2 years ago

Read 2 more answers

An object travels along a right triangle with sides 3 m, 4 m, and 5 m. It returns back to the initial position.

stiv31 [10]

When it travels 3m ,4m and 5m it means 12m is right answer.

3 0

3 years ago

Tourists covered 255 km for a 4-hour ride by car and a 7-hour ride by train. what is the speed of the train, if it is 5 km/h gre

LekaFEV [45]

The distance covered by car is equal to (assuming it is moving by uniform motion) the product between the car's speed and the time of the car ride, 4 h:
$S_c = v_c t_c$
where
$v_c$ is the car's speed
$t_c = 4 h$ is the duration of the car ride

Similarly, the distance covered by train is equal to the product between the train's speed and the duration of the train ride, 7 h:
$S_t = v_t t_t$

The total distance covered is S=255 km, which is the sum of the distances covered by car and train:
$S=255 km = S_c + S_t$
which becomes
$255 = 4 v_c + 7 v_t$ (1)
we also know that the train speed is 5 km/h greater than the car's speed:
$v_t = 5 + v_c$ (2)

If we put (2) into (1), we find
$255 = 4v_c + 7(5+v_c)$
and if we solve it, we find
$v_c = 20 km/h$
$v_t = 25 km/h$

So, the car speed is 20 km/h and the train speed is 25 km/h.

4 0

3 years ago

Force due to gravity. While JWST is in orbit, the Earth will be at a distance of 1.494 x 109 m from the telescope, and the Sun w

Alona [7]

Answer:

the gravitational force JWST will feel from the Sun is 37.785 Newton { leftward }

Explanation:

Given that;

distance of earth from the telescope r1 = 1.494 × 10⁹ m

and the Sun will be 1.49598 × 10¹¹ m further away

so r2 = r1 + 1.49598 × 10¹¹

r2 = 1.494 × 10⁹ m + 1.49598 × 10¹¹ m = 1.51092 × 10¹¹ m

mass of sun Ms = 1.9884 × 10³⁰ kg

mass of earth Me = 5.945× 10²⁴ kg

mass of JWST Mj = 6500 kg

What is the gravitational force JWST will feel from the Sun (strength and direction)?

the gravitational force of the sun will be attractive based on Newton law of gravitational force; so

Fjs = GMjMs / r2²

constant G = 6.674 × 10⁻¹¹ Nm²/kg²

Force on the JWST by the sun will be;

Fjs = GMjMs / r2² { leftward}

we substitute

Fjs = [(6.674 × 10⁻¹¹ Nm²/kg²)(6500 kg )(1.9884 × 10³⁰ kg)] / (1.51092 × 10¹¹ m)²

= (8.62587804 × 10²³) / ( 2.28287925 × 10²² )

= 37.785 Newton { leftward }

Therefore, the gravitational force JWST will feel from the Sun is 37.785 Newton { leftward }

7 0

2 years ago