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Vladimir [108]
4 years ago
11

What velocity must a car with a mass of 1110 kg have in order to have the same momentum as a 2280 kg pickup truck traveling at 2

4 m/s to the east? Answer in units of m/s.
Physics
1 answer:
shutvik [7]4 years ago
6 0

Answer:

49.3 m/s

Explanation:

The momentum is defined as the product of the object velocity and its mass.

So the momentum of the truck is

P_t = 2280 * 24 = 54720 kgm/s

For the car to have the same momentum, its speed must be

v_c = P_t/m_c = 54720 / 1110 = 49.3 m/s

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Only answer if you know all of them
lesya692 [45]
1)
The connections between neurons in the retina, specifically the connections referred to as “lateral inhibition,” help us see which of the following better?

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B) Faces
<span>C) Colors

2)
</span>Improving the contrast of an image (making the dark regions darker and the light regions lighter) helps us to identify:

<em><u>A) The edges of objects</u></em>
B) The center of objects
<span>C) The color of an object
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3)
What assumption does our visual system make in order to see curved surfaces (domes, holes)?

<em><u>A) Light comes from above</u></em>
B) Curved surfaces are always evenly lit
<span>C) Curved surfaces are always easy to see, no assumptions are made
</span>
4)
Which part of the face does our brain pay the most attention to?

<u><em>A) Eyes and mouth</em></u>
B) Eyes and ears
<span>C) Eyes and chin
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5)
If all these assumptions sometimes lead to mistakes, for example in these optical illusions, why do we make them?

A) It helps us see things faster
B) It helps us see things correctly
C) It helps us pay attention to what's important
<span><em><u>D) All of the above
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Hope that helps :)
*the correct answers are bolded, italicized, and underlined.*
4 0
3 years ago
What are the dark spots and light areas on the moon called
Tems11 [23]
The dark spots on the moon are the craters and there light spots are made up of the moon dust technically called regolith
5 0
4 years ago
The angular position of objects as a function of time is given, where a, b, and care constants. In which of these cases is the a
GalinKa [24]

Answer:

Explanation:

The options is not well presented

This are the options

A. θ = at³ + b

B. θ = at² + bt + c

C. θ = at² — b

D. θ = Sin(at)

So, we want to prove which of the following option have a constant angular acceleration I.e. does not depend on time

Now,

Angular acceleration can be determine using.

α = d²θ / dt²

α = θ''(t)

So, second deferential of each θ(t) will give the angular acceleration

A. θ = at³ + b

dθ/dt = 3at² + 0 = 3at²

d²θ/dt² = 6at

α = d²θ/dt² = 6at

The angular acceleration here still depend on time

B. θ = at² + bt + c

dθ/dt = 2at + b + 0 = 2at + b

d²θ/dt² = 2a + 0 = 2a

α = d²θ/dt² = 2a

Then, the angular acceleration here is constant is "a" is a constant and the angular acceleration is independent on time.

C. θ = at² —b

dθ/dt = 2at — 0 = 2at

d²θ/dt² = 2a

α = d²θ/dt² = 2a

Same as above in B. The angular acceleration here is constant is "a" is a constant and the angular acceleration is independent on time.

D. θ = Sin(at)

dθ/dt = aCos(at)

d²θ/dt² = —a²Sin(at) = —a²θ

α = d²θ/dt² = -a²θ

Since θ is not a constant, then, the angular acceleration is dependent on time and angular displacement

So,

The answer is B and C

4 0
3 years ago
Small fragments of orbiting bodies that have fallen on Earth's surface are known as
kolezko [41]
There are known as B. meteorites: asteroids are much larger.  
6 0
3 years ago
A circular conducting loop with a radius of 0.10 m and a small gap filled with a 10.0 ȍresistor is oriented in the xy-plane. If
lidiya [134]

Answer:

The magnitude of the current is 5.45 mA.

Explanation:

Given that,

Resistance = 10.0 ohm

Radius = 0.10 m

Magnetic field = 1.0 T

Angle = 30°

Increase magnetic field = 7.0 T

Time t = 3.0 s

Number of turns = 1

We need to calculate the initial flux

Using formula of flux

\phi=NB_{1}A\cos\theta

Put the value into the formula

\phi=1\times1.0\times\pi\times(0.10)^2\times\cos30^{\circ}

\phi=1\times1.0\times\pi\times(0.10)^2\times\dfrac{\sqrt{3}}{2}

\phi=0.027\ wb

We need to calculate the final flux

\phi=1\times7.0\times\pi\times(0.10)^2\times\dfrac{\sqrt{3}}{2}

\phi=0.1904\ wb

We need to calculate the induced emf

Using formula of emf

\epsilon=\dfrac{\phi_{f}-\phi_{i}}{t}

Put the value into the formula

\epsilon=\dfrac{0.1904-0.027}{3.0}

\epsilon=0.0545\ V

We need to calculate the current

Using formula of current

I=\dfrac{\epsilon}{R}

Put the value into the formula

I=\dfrac{0.0545}{10.0}

I=5.45\ mA

Hence, The magnitude of the current is 5.45 mA.

4 0
3 years ago
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