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almond37 [142]
3 years ago
14

Special relativity can be used to study an object in which frame of reference? A. A frame of reference that has no change in vel

ocity B. A frame of reference that is moving in a circle C. A frame of reference that has a constant, positive acceleration D. A frame of reference that has a constant, negative acceleratio
Physics
2 answers:
4vir4ik [10]3 years ago
5 0

Answer:

Explanation:d

Georgia [21]3 years ago
4 0

Answer:

gangster

Explanation:

sorrynsjajshdnqmanshsha

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A spearfisher stands in shallow water and sees a fish a few feet in front of her. She throws her spear directly toward the posit
DENIUS [597]

Answer:

the spear will end up above the fish relative to the actual position of the fish.

Explanation:

due to refraction of light coming from the fish the fish will appear slightly above from its real position

So due to this refraction the spearfisher will throw the spear directly at the image of the fish due to which it will not reach the position of fish but it will reach the position above the fish.

So here we can say that the spear will end up above the fish relative to the actual position of the fish

5 0
3 years ago
A delivery truck leaves a warehouse and travels 3.20 km east. The truck makes a right turn and travels 2.45 km south to arrive a
boyakko [2]

Explanation:

It is given that,

Displacement of the delivery truck, d_1=3.2\ km (due east)

Then the truck moves, d_2=2.45\ km (due south)

Let d is the magnitude of the truck’s displacement from the warehouse. The net displacement is given by :

d=\sqrt{d_1^2+d_2^2}

d=\sqrt{3.2^2+2.45^2}

d = 4.03 km

Let \theta is the direction of the truck’s displacement from the warehouse from south of east.

\theta=tan^{-1}(\dfrac{2.45}{3.2})

\theta=37.43^{\circ}

So, the magnitude and direction of the truck’s displacement from the warehouse is 4.03 km, 37.4° south of east.

8 0
3 years ago
Question is down below need rply fast
My name is Ann [436]

The X and Y components are as follows;

1. X = 35 * cos 57 = 19. 1m/s; Y = 35 * sin 57 = 29.4 m/s

2. X = 12 * -cos 34  = -10 m/s; Y = 12 * -sin 34 = -6.7 m/s

3. X = 8 * -cos 90  = 0 m/s; Y = 12 -sin 90 = -8 m/s

4. X = 20 * cos 75 = 5. 2m/s; Y = 20 * (-sin 75) =  -19.3 m/s

<h3>What are the horizontal and vertical components of the vectors?</h3>

The horizontal and vertical components of the velocities are given as follows:

  • Horizontal component, X = x cos θ
  • Vertical component, Y = y sin θ

1. 35 m/s at  57° from x-axis

X = 35 * cos 57 = 19. 1m/s

Y = 35 * sin 57 = 29.4 m/s

2. 12m/s at 34° S of W

X = 12 * -cos 34  = -10 m/s

Y = 12 * -sin 34 = -6.7 m/s

3. 8 m/s at South

X = 8 * -cos 90  = 0 m/s

Y = 12 -sin 90 = -8 m/s

4. 20 m/s at  275° from x-axis

X = 20 * cos 75 = 5. 2m/s

Y = 20 * (-sin 75) =  -19.3 m/s

In conclusion, the X and Y components are found by taking cosines and sine of the angles.

Learn more about horizontal and vertical components at: brainly.com/question/26446720

#SPJ1

8 0
1 year ago
Two long, straight wires are parallel and 16 cm apart. One carries a current of 2.9 A, the other a current of 5.7 A. (a) If the
Mnenie [13.5K]

Explanation:

Given that,

Distance between two long wires, d = 16 cm = 0.16 m

Current in one wire, I_1=2.9\ A

Current in wire 2, I_2=5.7\ A    

The magnetic force per unit length of one wire on the other is given by the following expression as :

\dfrac{F}{l}=\dfrac{\mu_o I_1I_2}{2\pi d}

\dfrac{F}{l}=\dfrac{4\pi \times 10^{-7}\times 2.9\times 5.7}{2\pi \times 0.16}

\dfrac{F}{l}=2.06\times 10^{-5}\ N/m

The current is flowing in opposite direction, the magnetic force acting on it is repulsive. Hence, this is the required solution.

5 0
3 years ago
Which statement is true about the gravitational force?
topjm [15]

Hello,

I think that A is the right one.

6 0
3 years ago
Read 2 more answers
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