Ask question

Ask question

All categories

2 years ago

7

Suzie skydiver, who weighs 500 n, reaches terminal velocity of 90 km/h. the air resistance on suzie is then

2 answers:

Gekata [30.6K]2 years ago

8 0

Answer: Air resistance would be 625Newton

Explanation: For small object the air resistance is directly proportional to the velocity that is if you triple the Velocity the air resistance is also tripled.

But for large object like ball and human the air resistance is proportional to the square of the Velocity that is V². So in this case the velocity given is 90km/h which is also 25 m/s.

So the air resistance on Suzie would be the square of the Velocity which is 25² = 625N

masha68 [24]2 years ago

7 0

<span>Weight of the skydiver m = 500 N
Terminal velocity V = 90 km/h
Here the weight of the person acts as the force, so based on the Newton's third law the applied is the force what we but in the opposite direction making the resistance. So the air resistance exerted on Suzie will be her weight that is 500N</span>

You might be interested in

A block spring system oscillates on a frictionless surface with an amplitude of 10\text{ cm}10 cm and has an energy of 2.5 \text

antoniya [11.8K]

Answer:

The energy of the system is 15 J.

Explanation:

Given that,

Energy E = 2.5 J

Amplitude = 10 cm

We need to calculate the spring constant

Using formula of mechanical energy of the system

$E=\dfrac{1}{2}kA^2$

Put the value into the formula

$2.5=\dfrac{1}{2}k\times(10\times10^{-2})^2$

$k=\dfrac{2.5\times2}{(10\times10^{-2})^2}$

$k=500\ N/m$

If the block is replaced by a block with twice the mass of the original block

Amplitude = 6 cm

We need to calculate the energy

Using formula of mechanical energy

$E=\dfrac{1}{2}kA^2$

Put the value into the formula

$E=\dfrac{1}{2}\times500\times(6\times10^{-2})$

$E=15\ J$

Hence, The energy of the system is 15 J.

8 0

2 years ago

A cow’s mass is 410 kg and a car’s mass is 565 kg. What is the difference between their weights?

solmaris [256]

B. 1520 is the difference between their weights.

5 0

2 years ago

Centripetal force Fc acts on a car going around a curve. If the speed of the car were twice as great, the magnitude of the centr

kondaur [170]

Answer:

We need 4 times more force to keep the car in circular motion if the velocity gets double.

Explanation:

Lets take the mass of the car = m

The radius of the arc = r

$F=\frac{m\times v^2}{r}$

Given that speed of the car gets double ,v' = 2 v

Then the force on the car = F'

$F'=\frac{m\times v'^2}{r}$ ( radius of the arc is constant)

$F'=\frac{m\times (2v)^2}{r}$

$F'=4\times \frac{m\times v^2}{r}$

We know that $F=\frac{m\times v^2}{r}$

Therefore F' = 4 F

So we can say that we need 4 times more force to keep the car in circular motion if the velocity gets double.

6 0

2 years ago

The electric field at a point 2.8 cm from a small object points toward the object with a strength of 180,000 N/C. What is the ob

givi [52]

Answer:

Charge, $Q=1.56\times 10^{-8}\ C$

Explanation:

It is given that,

Electric field strength, E = 180000 N/C

Distance from a small object, r = 2.8 cm = 0.028 m

Electric field at a point is given by :

$E=\dfrac{kQ}{r^2}$

Q is the charge on an object

$Q=\dfrac{Er^2}{k}$

$Q=\dfrac{180000\ N/C\times (0.028\ m)^2}{9\times 10^9\ Nm^2/C^2}$

$Q=1.56\times 10^{-8}\ C$

So, the charge on the object is $1.56\times 10^{-8}\ C$ . Hence, this is the required solution.

7 0

2 years ago

A boy can accelerate at 1.00 m/s 2 over a short distance. If the were to take an energy pill and suddenly have the ability to ac

VikaD [51]

We know that:
F = ma,
so
F/a = constant
F₁/a₁ = F₂/a₂
F₂ = F₁ x 5.6 / 1
F₂ = 5.6F₁

If F₁ = 45 N,
F₂ = 5.6 x 45
= 252 N

4 0

2 years ago