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3 years ago

11

If you wanted to move an electron from the positive to the negative terminal of the battery, how much work W would you need to d

o on the electron?
Enter your answer numerically in joules.

1 answer:

brilliants [131]3 years ago

8 0

a lot of electric so that the battery can work

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Which of the following scientists did not contribute scientific discoveries to physical science? A) Isaac Newton B) Marie Curie

lianna [129]

Answer:

It is Gregor Mendel because he was a Catholic Augustinian monk and naturalist

Explanation:

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3 years ago

a boy is standing 4 meter from a plane mirror how far and in what distance must te move so that he will be 4 meter from his imag

Alona [7]

Answer:

2 meters towards the mirror.

Explanation:

In a plane mirror the image distance is equal to the object distance. Therefore, by moving 2 meters towards the mirror, the boy reduces the distance between him and the mirror to two meters which is the object distance. The image distance is also 2 meters. add the two distances you will get four meters.

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3 years ago

An analogy makes a comparison between objects based on their similar qualities. Cassidy wanted to create an analogy for the moti

user100 [1]

The three phases of matter differ in properties just because of the proximity of their molecules. The solid phase is the most organized of all. Its atoms are compactly arranged together and has the strongest intermolecular forces to keep them together. This is why they have a definite shape and volume. The liquid phase have molecules that are far away from each other, but not as far as that of the gas phase. The liquid and gas phases can be lumped into one group called fluids because they have the same property - they take the shape and volume of their container.

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3 years ago

Read 2 more answers

Attempt 2 You have been called to testify as an expert witness in a trial involving a head-on collision. Car A weighs 15151515 l

Natasha_Volkova [10]

Answer:

v = 28.98 ft / s

Explanation:

For this problem we must solve it in parts, let's start by looking for the speed of the two cars after the collision

In the exercise they indicate the weight of each car

Wₐ = 1500 lb

W_b = 1125 lb

Car B's velocity from v_b = 42.0 mph westward, car A travels east

let's find the mass of the vehicles

W = mg

m = W / g

mₐ = Wₐ / g

m_b = W_b / g

mₐ = 1500/32 = 46.875 slug

m_b = 125/32 = 35,156 slug

Let's reduce to the english system

v_b = 42.0 mph (5280 foot / 1 mile) (1h / 3600s) = 61.6 ft / s

We define a system formed by the two vehicles, so that the forces during the crash have been internal and the moment is preserved

we assume the direction to the east (right) positive

initial instant. Before the crash

p₀ = mₐ v₀ₐ - m_b v_{ob}

final instant. Right after the crash

p_f = (mₐ + m_b) v

the moment is preserved

p₀ = p_f

mₐ v₀ₐ - m_b v_{ob} = (mₐ + m_b) v

v = $\frac{ m_a \ v_{oa} - m_b \ v_{ob} }{ m_a +m_b}$

we substitute the values

v = $\frac{ 46.875}{82.03} \ v_{oa} - \frac{35.156}{82.03} \ 61.6$

v = 0.559 v₀ₐ - 26.40 (1)

Now as the two vehicles united we can use the relationship between work and kinetic energy

the total mass is

M = mₐ + m_b

M = 46,875 + 35,156 = 82,031 slug

starting point. Jsto after the crash

K₀ = ½ M v²

final point. When they stop

K_f = 0

The work is

W = - fr x

the negative sign is because the friction forces are always opposite to the displacement

Let's write Newton's second law

Axis y

N-W = 0

N = W

the friction force has the expression

fr = μ N

we substitute

-μ W x = Kf - Ko

-μ W x = 0 - ½ (W / g) v²

v² = 2 μ g x

v = $\sqrt{ 2 \ 0.750 \ 32 \ 17.5}$ Ra (2 0.750 32 17.5

v = 28.98 ft / s

3 0

2 years ago

HELP!!! 30 POINTS+BRAINLIEST!!!!!QUICK!!

tia_tia [17]

Answer:

A:7.2 $ms^{-1}$

B:14.25 $ms^{-1}$

C:1.45 $sec$

D:10.3 $m$

E:2.9 $sec$

F:20.88 $m$

Explanation:

Let $v$ be the velocity and $\alpha$ be the angle between the velocity and ground.

Question A:

Horizontal component of velocity is given by $vcos(\alpha )$ .

So,horizontal component is $16\times cos(63)=16\times 0.45=7.2ms^{-1}$

Question B:

Vertical component of velocity is given by $vsin(\alpha )$ .

So,vertical component is $16\times sin(63)=16\times 0.89=14.25ms^{-1}$

Question C:

Time required is given by $\frac{\text{vertical component of velocity}}{g}}=\frac{14.25}{9.8}=1.45 seconds$

Question D:

Maximum height is given by $\frac{\text{vertical component of velocity}^{2}}{2g}}=\frac{203.06}{19.6}=10.3m$

Question E:

Time of flight is twice the time required to reach maximum height= $2\times 1.45=2.9 seconds$ .

Question F:

The distance between the player and ball after landing is called range and is given by $\text{horizontal component of velocity}\times \text{time of flight}=$ $7.2\times 2.9=20.88m$

8 0

3 years ago