I appreciate the help on this physics question

The drag force pushes opposite your motion as you ride a bicycle. If you double your speed, what happens to the drag force?

timurjin [86]

Answer: The drag force goes up by a factor of 4

Explanation:

The Drag Force equation is:

$F_{D}=\frac{1}{2}C_{D}\rho A_{D}V^{2}$ (1)

Where:

$F_{D}$ is the Drag Force

$C_{D}$ is the Drag coefficient, which depends on the material

$\rho$ is the density of the fluid where the bicycle is moving (air in this case)

$A_{D}$ is the transversal area of the body or object

$V$ the bicycle's velocity

Now, if we assume $C_{D}$ , $\rho$ and $A_{D}$ do not change, we can rewrite (1) as:

$F_{D}=C.V^{2}$ (2)

Where $C$ groups all these coefficients.

So, if we have a new velocity $V_{n}$ , which is the double of the former velocity:

$V_{n}=2V$ (3)

Equation (2) is written as:

$F_{D}=C.V_{n}^{2}=C.(2V)^{2}$

$F_{D}=4CV^{2}$ (4)

Comparing (2) and (4) we can conclude the Drag force is four times greater when the speed is doubled.

7 0

2 years ago

In a large restaurant an average 60% customers ask for water with their meal. A random sample of 10 customers is selected. Find

Gekata [30.6K]

Answer:

a) $P(6)=0.25$

b) $p(x$

c) $p(x\geq3)=0.9878$

d) $\sigma=\sqrt{2.4}=1.5492$

Explanation:

From the question we are told that:

Population percentage $p_\%=\60%$

Sample size $n=10$

Let x =customers ask for water

Let y =customers dose not ask for water with their meal

Generally the equation for y is mathematically given by

$y=1-p_\%\\y=1-0.60\\y=0.40$

Generally the equation for pmf p(x) is mathematically given by

$P(x)=10C_x (0..6)^x(0.4)^{10-x}$

a)

Generally the probability that exactly 6 ask for water is mathematically given by

$P(x)6=10C_6 (0..6)^6(0.4)^{10-6}$

$P(6)=0.25$

b)

Generally the probability that less than 9 ask for water with meal is mathematically given by

$p(xg)$

$p(x$

c)

Generally the probability that at least 3 ask for water with meal is mathematically given by

$p(x\geq3)=1-p(x$

$p(x\geq3)=1-[p(0)+p(1)+p(2)]$

$p(x\geq3)=1-[0.00001+0.0015+0.0106]$

$p(x\geq3)=1-[0.0122]$

$p(x\geq3)=0.9878$

d)

Generally the mean and standard deviation of sample size is mathematically given by

Mean

$\=x=np=10(0.6)=6$

Standard deviation

$v(x)=npq=10(0.6)(0.4)=2.4$

$\sigma=\sqrt{2.4}=1.5492$

4 0

2 years ago

SP: Calculate the moment

ipn [44]

Answer:

Moment of the force is 20 N-m.

Explanation:

Given:

Force exerted by the person is, $F=80\ N$

Distance of application of force from the point about which moment is needed is, $d=25\ cm=\frac{25}{100}\ m=0.25\ m$

Now, we know that, moment of a force 'F' about a point at a perpendicular distance of 'd' from the same point is given as the product of the force and the perpendicular distance.

Therefore, the moment of the force about the end of the claw hammer is given as:

$M=F\times d\\\\M=(80\ N)(0.25\ m)\\\\M=20\textrm{ N-m}$

Hence, the moment of the force exerted by the person about the end of the claw hammer is 20 N-m.

6 0

2 years ago

A 1000-kg car is moving at 30 m/s around a horizontal unbanked curve whose diameter is 0.20 km. What is the magnitude of the fri

omeli [17]

Answer:

4500 N

Explanation:

When a body is moving in a circular motion it will feel an acceleration directed towards the center of the circle, this acceleration is:

a = v^2/r

where v is the velocity of the body and r is the radius of the circumference:

Therefore, a body with mass m, will feel a force f:

f = m v^2/r

Therefore we need another force to keep the body(car) from sliding, this will be given by friction, remember that friction force is given a the normal times a constant of friction mu, that is:

fs = μN = μmg

The car will not slide if f = fs, i.e.

fs = μmg = m v^2/r

That is, the magnitude of the friction force must be (at least) equal to the force due to the centripetal acceleration

fs = (1000 kg) * (30m/s)^2 / (200 m) = 4500 N

7 0

3 years ago

Read 2 more answers

Please help have due very soon?thank you

Andru [333]

The answer is :78 I think

8 0

2 years ago