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3 years ago

I need help with this science question, 50 points!

Physics

1 answer:

Gennadij [26K]3 years ago

7 0

Answer:

an open circuit (A battery, bulb, wire,switch)

Explanation:

the battery supplies the power the wire carries the current from the battery to the bulb but the circuit is not completed if the switch is not connected (Switch )

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The variables for this experiment include mass, volume, and the materials in the various balls and their densities. In Part III,

zmey [24]

Answer:

volume and density

Explanation:

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2 years ago

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ludmilkaskok [199]

Answer: W represents an element Y represents a mixture

Explanation: I just took a quiz on this and got it right. Good Luck

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3 years ago

How many joules of gravitational potential energy does a person have after they went up stairs that are 10 meters tall? Assume t

Sonja [21]

Answer:

P = 5880 J

Explanation:

Given that,

The mass of a person, m = 60 kg

The height of the stairs, h = 10 m

We need to find the gravitational potential energy of the person. The formula is as follows :

$P=mgh$

Substitute all the values,

$P=60\times 9.8\times 10\\\\P=5880\ J$

So, the required gravitational potential energy is equal to 5880 J.

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3 years ago

If a projectile is launched vertically from the Earth with a speed equal to the escape speed, how high above the earth's surface

Alecsey [184]

Answer: h = 3R

Explanation:

Using the law of conservation of energy,

Total energy at the beginning of the launch would be equal to total energy at any point.

kinetic energy + gravitational potential energy = constant

Initial energy of the projectile = $\frac{1}{2}mv_e^2-\frac{GMm}{R}$ ... (1)

where R is the radius of the Earth, M is the mass ofthe Earth, m is the mass of the projectile.

escape velocity, $v_e=\sqrt{\frac{2GM}{R}}$

Total energy at height h above the Earth where speed of the projectile is half the escape velocity:

$\frac{1}{2}m(v_e/2)^2-G\frac{Mm}{R+h}$ ...(2)

(1)=(2)

⇒ $\frac{1}{2}m(v_e/2)^2-G\frac{Mm}{R+h} = \frac{1}{2}mv_e^2-\frac{GMm}{R}$

⇒ $G\frac{Mm}{R}-G\frac{Mm}{R+h}= \frac{1}{2}m \frac{3}{4}v_e^2 =\frac{1}{2}m \frac{3}{4} (\sqrt{\frac{2GM}{R}})^2$

⇒ $\frac{GM(R+h-R)}{R(R+h)} = \frac{3}{4}(\frac{GM}{R})$

⇒ $\frac{h}{R+h} = \frac{3}{4}$

⇒h = 3R

Thus, at height equal to thrice radius of Earth, the speed of the projectile would reduce to half of escape velocity.

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3 years ago

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find the coefficient of kinetic friction for a 10 kg box being dragged steadily across the surface with a force of 2.0 Newtons

Mazyrski [523]

Answer:

0.02

Explanation:

Force, $F=\mu_k N$ where N is normal reaction and coefficient of kinetic friction is $\mu_k$ . Also, N is equivalent to mg ie N=mg where m is mass of an object and g is acceleration due to gravity. Making $\mu$ the subject of the formula then

$\mu_k=\frac {F}{mg}$

Substituting F with 2 N, m with 10 kg and g with 9.81 then

$\mu_k=\frac {2}{10\times 9.81}=0.0203873598369\approx 0.02$

5 0

4 years ago