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3 years ago

Someone please help me will give BRAILIEST!!!!!

Physics

1 answer:

ratelena [41]3 years ago

4 0

The speed increases, because as the angle increases (the wing slants up more steeply), the air has to go farther to get over the wing.

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The linear impulse delivered by the hit of a boxer is 202 N · s during the 0.244 s of contact. What is the magnitude of the aver

zlopas [31]

Answer: Magnitude of the average force exerted on the glove by the other boxer is 827.86 N (approximately 828 N).

Explanation: Impulse is defined as the force acting on an object for a short period or interval of time.

Mathematically it is given by the relation:

Impulse = Force $\times$ Time

According to the numerical values given in the question, I = 202 Ns and T = 0.244 s

So, Force F = $\frac{Impulse}{Time}$ = $\frac{202}{0.244}$ = 827.86 N

Magnitude of the average force exerted on the glove by the other boxer is 827.86 N (approximately 828 N).

7 0

3 years ago

Which is not a step of the scientific method?

Dahasolnce [82]

The six steps of the scientific are:
1. State the question
2. Conduct research
3. Create a hypothesis
4. Perform the experiment
5. Analyze the data
6. Conclusion

So D. would be the correct answer, even though communicating the results could possibly be a step if it's required.

4 0

3 years ago

What force is acting on the rainwater in the model?

Vikki [24]

The force is gravitational because when something is falling is call gravitational

8 0

3 years ago

Read 2 more answers

to measure the static friction coefficient between a block and a vertical wall, a spring is attached to the block, is pushed on

Stolb23 [73]

Answer:

μ = mg/kx

Explanation:

Since the bock does not slip, the frictional force equals the weight of the block. So, F = mg. Now, the frictional force, F = μN where μ = coefficient of static friction and N = Normal force.

Now, the normal force equals the spring force F' = kx where k = spring constant and x = compression of spring.

N = F' = kx

So, F = μN = μkx

μkx = mg

So, μ = mg/kx

8 0

3 years ago

The planet Mars has a mass of 6.1 × 1023 kg and radius of 3.4 × 106 m. What is the acceleration of an object in free fall near t

Oksanka [162]

The acceleration due to gravity of Mars is $3.5\ m / s^{2}$

<u>Explanation:</u>

As per universal law of gravity, the gravitational force is directly proportional to the product of masses and inversely proportional to the square of the distance between them. But in the present case, the gravity need to be determined between Mars and the object on Mars. Since the mass of Mars is greater than the mass of any object. Thus,

$\text {Gravitational force of planet}=\frac{G M m}{R^{2}}$

Here, G is the gravitational constant, R is the radius of Mars and M, m is the mass of Mars and the object respectively..

Also, according to Newton’s second law of motion, the acceleration of any object will be equal to the ratio of force exerted on it to the mass of the object.

So in order to determine the acceleration due to gravity of Mars, divide the gravitational force of Mars by mass of object on the surface of Mars.

$Acceleration\ due\ to\ gravity\ of\ mars =\frac{\text {Gravitational force of Mars}}{m \text { of object }}$

$Acceleration\ due\ to\ gravity\ of\ mars =\frac{G M m}{R^{2} \times m}=\frac{G M}{R^{2}}$

$\text { Acceleration due to gravity of mars }=\frac{6.67259 \times 10^{-11} \times 6.1 \times 10^{23}}{3.4 \times 3.4 \times 10^{12}}=\frac{40.703 \times 10^{12}}{11.56 \times 10^{12}}$

$\text { Acceleration }=3.5\ \mathrm{m} / \mathrm{s}^{2}$

3 0

3 years ago