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gavmur [86]
3 years ago
7

A car traveling south is 200 kilometers from its starting point after 2 hours. What is the average velocity of the car?

Physics
2 answers:
MakcuM [25]3 years ago
6 0
The velocity of the car would be 100 kilometer per hour.
MrRissso [65]3 years ago
3 0

The answer is a. 100 kilometers /hour south

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6 latter word and it has a s and a I and it has mass (9.______________liquids, and gases all have mass.)
Firdavs [7]

Answer:

Solids

Explanation:

Solids, liquids, and gases all have mass.

7 0
2 years ago
What does the zigzag line in the circuit diagram represent? A. a source of electrical energy B. an electrical conductor C. an el
lora16 [44]
C. an electrical resistor 
8 0
3 years ago
Read 2 more answers
[High Dive) above a pool of water. According to the announcer, the divers enter the water at a speed of 56 mi/h (25 m/s). (Air r
blsea [12.9K]

Answer:

(a) The announcer's claim is incorrect because the divers enter at a speed of 20.4 and not 25 m/s as announced

(b)  it’s possible for a diver to enter the water with the velocity of 25 m/s if he has initial velocity of 14.4 m/s. The upward initial velocity can’t be physically attained

Explanation:

(a)

To find the final velocity V_{f} for an object traveling distance h taking the initial vertical component of velocity as V_{i} the kinematics equation is written as

V_{f}^{2}=V_{i}^{2}+2ah where a is acceleration

Substituting g for a where g is gravitational force value taken as 9.81

V_{f}^{2}=V_{i}^{2}+2gh

Since the initial velocity is zero, we can solve for final velocity by substituting figures, note that 70 ft is 21.3 m for h

V_{f}=\sqrt {(2gh)}= V_{f}=\sqrt {(2*9.81*21.3)}= 20.44275

Therefore, the divers enter with a speed of 20.4 m/s

The announcer's claim is incorrect because the divers enter at a speed of 20.4 and not 25 m/s as announced

(b)

The divers can enter water with a velocity of 25 m/s only if they have some initial velocity. Using the kinematic equation

V_{f}^{2}=V_{i}^{2}+2gh

Since we have final velocity of 25 m/s

V_{i}^{2}=2gh-V_{f}^{2}

V_{i}=\sqrt{(V_{f}^{2}-2gh)}

V_{i}=\sqrt{(25^{2}-2*9.81*21.3)}= 14.390761 m/s

Therefore, it’s possible for a diver to enter the water with the velocity of 25 m/5 if he has initial velocity of 14.4 m/s

In conclusion, the upward initial velocity can’t be physically attained

3 0
3 years ago
What is the force exerted on a charge of 2.5 µC moving perpendicular through a magnetic field of 3.0 × 102 T with a velocity of
OverLord2011 [107]

Given:

B = 3 \times 10^{2} T

V=5 \times 10^{3} \frac{m}{s}

q = 2.5 × 10^{-6} C

α = 90

To find:

Force = ?

Formula used:

Force on the moving charge is given by,

F = q V B sin α

Where F = force exerted on moving charge

V = velocity of charge

q = charge

α = angle between direction of V and B

Solution:

F = q V B sin α

Where F = force exerted on moving charge

V = velocity of charge

q = charge

α = angle between direction of V and B

F = 2.5 \times 10^{-6} \times 3 \times 10^{2}  \times 5 \times 10^{3}

F = 37.5 × 10^{-1}

F = 3.75 Newton

Thus, the force acting on the moving charge is 3.75 Newton.

8 0
3 years ago
Read 2 more answers
In the macroscopic world, you know that you can hear but cannot see around corners. Under what conditions does light bend around
omeli [17]

Answer:

a much larger slit, the phenomenon of Sound diffraction that slits for light.

this is a series of equally spaced lines giving a diffraction envelope

Explanation:

The diffraction phenomenon is described by the expression

    d sin θ = m λ

Where d is the distance of the slit, m the order of diffraction that is an integer and λ the wavelength.

 

For train the diffraction phenomenon, the d / Lam ratio is decisive if this relation of the gap separation in much greater than the wavelength does not reduce the diffraction phenomenon but the phenomena of geometric optics.

The wavelength range for visible light is 4 10⁻⁷ m to 7 10⁻⁷ m. The wavelength range for sound is 17 m to 1.7 10⁻² m. Therefore, with a much larger slit, the phenomenon of Sound diffraction that slits for light.

When we add a second slit we have the diffraction of each one separated by the distance between them, when the integrals are made we arrive at the result of the interference phenomenon, a this is a series of equally spaced lines giving a diffraction envelope

When I separate the distance between the two slits a lot, the time comes when we see two individual diffraction patterns

6 0
3 years ago
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