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Mars2501 [29]
3 years ago
12

If it actually hits the ground with a speed of 8.50 m/s , what is the magnitude of the average force of air resistance exerted o

n it?
Physics
1 answer:
kkurt [141]3 years ago
8 0
Well it will probably never hit the ground try to look over your answer and go from there I hope I helped
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Darius' boat sails into the harbor with a speed of 80m/s. After 20 seconds, Darius' boat has come to a stop at the dock. What is
Novosadov [1.4K]

Answer:

The boat's acceleration is 4 m/s²

Explanation:

This question seeks to test the knowledge of acceleration and how to calculate acceleration when the speed is provided. Hence, the formula for acceleration would be used here.

Acceleration (m/s²) = speed (in m/s) ÷ time (in seconds)

Acceleration = 80 m/s ÷ 20 secs

Acceleration = 4 m/s² or 4 ms⁻²

The boat's acceleration is 4 m/s²

4 0
2 years ago
Answer the following questions
zaharov [31]

Answer:

9 - 10N to the left

10 - There is no change on the object

Explanation:

Can I have brainliest answer pls?

7 0
2 years ago
Read 2 more answers
Is India a rich country?
Norma-Jean [14]

Explanation:

India. Total wealth: $8.9 trillion | Wealth per capita: $6,440 | India, which is the fifth-largest economy in the world, is home to 3,57,000 HNWIs and 128 billionaires.

4 0
2 years ago
Read 2 more answers
A kite 100 ft above the ground moves horizontally at a speed of 11 ft/s. At what rate is the angle (in radians) between the stri
frutty [35]

Answer:

-2.26×10^-4 radians

Explanation:

The solution involves a right angle triangle

Length is z while the horizontal is the height x

X^2+ 100^2=z^2

Taking the derivatives

2x(dx/dt)=Z^2(dz/dt)

Specific moments = Z= 200 ,X= 100sqrt3 and dx/dt= 11

dz/dt= 1100sqrt3/200 = 9.53

Sin a= 100/a

Taking derivatives in terms of t

Cos a(da/dt)=100/z^2 dz/dt

a= 30°

Cos (30°)da/dt= (-100/40000×9.5)

a= -2.26×10^-4radians

8 0
3 years ago
Read 2 more answers
A conductor carrying a current I = 16.5 A is directed along the positive x axis and perpendicular to a uniform magnetic field. A
Jet001 [13]

To solve this problem we will apply the concepts related to the Magnetic Force, this is given by the product between the current, the body length, the magnetic field and the angle between the force and the magnetic field, mathematically that is,

F = ILBsin \theta

Here,

I = Current

L = Length

B = Magnetic Field

\theta = Angle between Force and Magnetic Field

But \theta = 90\°

F = ILB

Rearranging to find the Magnetic Field,

B = \frac{F}{IL}

Here the force per unit length,

B = \frac{1}{I}\frac{F}{L}

Replacing with our values,

B = \frac{0.130N/m}{16.5}

B = 0.0078T

Therefore the magnitude of the magnetic field in the region through which the current passes is 0.0078T

6 0
3 years ago
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