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3 years ago

What factors effect the amount of momentum of an object has?

Physics

1 answer:

V125BC [204]3 years ago

6 0

Mass, speed and direction (if direction is important).

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An electric toy with a resistance of 2.50 Ω is operated by a 3.00-V battery. (a) What current does the toy draw? (b) Assuming th

Leno4ka [110]

Answer:

a) The current is i = 1.2 A

b) The charge is Q = 17280 C

c) The energy is E = 43200 J

Explanation:

a) The current is given by the ohm's law wich is:

i = V/R = 3/2.5 = 1.2 A

b) Since the charge is steady we can use the following equation to find the charge amount in that time:

i = Q/t

Q = t*i

Where t is in seconds, so we have 4h * 3600 = 14400 s

Q = 1.2*14400 = 17280 C

c) The energy is the power delivered to the toy multiplied by the time:

P = 1.2*2.5 = 3 W

E = P*t = 3*14400 = 43200 J

7 0

3 years ago

White moths were found in huge numbers in the city of Birmingham. These moths could be easily camouflaged by the trees. Because

Akimi4 [234]

A:The Population Will Decrease.

6 0

3 years ago

Read 2 more answers

A model rocket is launched straight upward from the surface of the Earth at a speed of 50 m/s. If air resistance is negligible,

romanna [79]

Answer:

The instantaneous velocity of the rocket the moment before it hits the ground is 50 m/s.

Explanation:

Given;

initial velocity of the rocket, u = 50 m/s

Determine the maximum height reached by the rocket.

at maximum height reached by the rocket, the final velocity, v = 0

v² = u² -2gh

0 = 50² - 2(9.8)h

19.6h = 2500

h = 2500 / 19.6

h = 127.55 m

At maximum height, the time to reach ground is given by;

h = ¹/₂gt²

$t = \sqrt{\frac{2h}{g} } \\\\t = \sqrt{\frac{2*127.55}{9.8} }\\\\ t = 5.1 \ s$

Before the rocket hits the ground the final velocity will be maximum;

v = u + gt

v = 0 + 9.8 x 5.1

v = 50 m/s

Therefore, the instantaneous velocity of the rocket the moment before it hits the ground is 50 m/s.

3 0

4 years ago

Aerobic exercise is exercise where oxygen is not present.

Alex17521 [72]

I say it is false that is the correct answer

4 0

4 years ago

What area must the plates of a capacitor be if they have a charge of 5.7uC and an electric field of 3.1 kV/mm between them? O 0.

kirill [66]

Answer:

Area of the plates of a capacitor, A = 0.208 m²

Explanation:

It is given that,

Charge on the parallel plate capacitor, $q = 5.7\ \mu C=5.7\times 10^{-6}\ C$

Electric field, E = 3.1 kV/mm = 3100000 V/m

The electric field of a parallel plates capacitor is given by :

$E=\dfrac{q}{A\epsilon_o}$

$A=\dfrac{q}{E\epsilon_o}$

$A=\dfrac{5.7\times 10^{-6}\ C}{3100000\ V/m\times 8.85\times 10^{-12}\ F/m}$

A = 0.208 m²

So, the area of the plates of a capacitor is 0.208 m². Hence, this is the required solution.

8 0

3 years ago