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Studentka2010 [4]
3 years ago
9

How does friction affect speed?

Physics
1 answer:
weeeeeb [17]3 years ago
7 0

Answer:

Friction is a force tends to oppose relative motion between bodies when they are in direct contact. Remember this friction does not oppose motion, it opposes relative motion.

Explanation:

You might be interested in
A cart of mass M = 2.40 kg can roll without friction on a level track. A light string draped over a light, frictionless pulley c
sergiy2304 [10]

Answer:

Part a)

a_{hanger} = g - \frac{T}{m}

Part b)

a_{cart} = \frac{T}{M}

Part c)

a = \frac{mg}{M + m}

Part d)

a = 1.35 m/s^2

Explanation:

Part a)

For hanger we know that it will have tension force upwards while it has downwards its weight so we will have

mg - T = ma

so we have

a_{hanger} = g - \frac{T}{m}

Part b)

now for car that is rolling on the floor the net force is given as

F = Ma

T = Ma

a_{cart} = \frac{T}{M}

Part c)

now we know that the cart and the hanger both are connected to each other

so they must have same acceleration

so we will have

T = Ma

mg - Ma = ma

a = \frac{mg}{M + m}

Part d)

now we know that

M = 2.40 kg

m = 0.50 kg

so we will have

a = \frac{0.50(9.81)}{2.40 + 0.50}

a = 1.35 m/s^2

6 0
3 years ago
You serve a tennis ball from a height of 1.80 m above the ground. The ball leaves your racket with a velocity of 18.0 m/s at an
Delicious77 [7]

Answer:

Yes, ball will clear the net

Explanation:

First we have to find the range of projectile motion.

Data given,

Ф = 7°

Initial velocity = 18 m/s

R = (V)^2.sin2Ф/g

Now by putting values

R = 7.99 m

Now for height

h = v^2.(sinФ)^2/2g

by putting values

h = 0.245 m

Since range is less than our distance (11.83 m) from net, so still it is not clear that ball will clear the net or not.

So, now from the maximum height, we have to calculate the horizontal distance of ball to net.

Now velocity in projectile motion is in two dimensions.

V(x) = 18 m/s

V(y) = 0 m/s (because at maximum height, ball will stop and then start again, so y-component of velocity will be 0 but since there will be no acceleration along x-axis, so V(x) will be 18 m/s)

Now, by formula S = V(y)t + (1/2)gt^2

we can calculate time which is required by the ball to reach net from the maximum height it has achieved.

Now, tricky part is to calculate S, because without it we can not calculate t.

So, by data given in question, we know that the ball is served at height of 1.8 m and it achieved the height of 0.245 m. But net is at height of 1.07 m.

So, the vertical distance downward, which ball will travel from maximum height to net will be

S = 1.8 + 0.245 - 1.07

S = 0.975 m

Since we know V(y) = 0 m/s

S = (1/2)gt^2

t = (2S/g)^(1/2)

t = 0.44 s

Now time for both vertical and horizontal distance are same,

So, for horizontal distance "D(x)"

D(x) = V(x) x t (Since, no acceleration along x axis, so we can use simple formula to calculate distance)

D(x) = 18 x 0.44

D(x) = 8.029 m

Now please notice that at maximum height, range was half, so at that point ball covered distance "a"

a = 3.99 m

From maximum height to net, as we calculated, ball covered

D(x) = 8.029 m

So, total distance covered by ball

a + D(x) = 3.99 + 8.029

a + D(x) = 12.024 m

which is more than your total distance from net which is 11.83 m. So, the ball will clear the net.

7 0
3 years ago
The graph of the relationship between the volume of a gas at constant temperature and its pressure is a(n)?
timurjin [86]
<span>When the difference between two results is larger than the estimates error, the result is</span>
5 0
3 years ago
A cylinder of radius R, length L, and mass M is released from rest on a slope inclined at angle θ. It is oriented to roll straig
inna [77]

Answer:

\mu_s=\frac{1}{3}\tan \theta

Explanation:

Let the minimum coefficient of static friction be \mu_s.

Given:

Mass of the cylinder = M

Radius of the cylinder = R

Length of the cylinder = L

Angle of inclination = \theta

Initial velocity of the cylinder (Released from rest) = 0

Since, the cylinder is translating and rolling down the incline, it has both translational and rotational motion. So, we need to consider the effect of moment of Inertia also.

We know that, for a rolling object, torque acting on it is given as the product of moment of inertia and its angular acceleration. So,

\tau =I\alpha

Now, angular acceleration is given as:

\alpha = \frac{a}{R}\\Where, a\rightarrow \textrm{linear acceleration of the cylinder}

Also, moment of inertia for a cylinder is given as:

I=\frac{MR^2}{2}

Therefore, the torque acting on the cylinder can be rewritten as:

\tau = \frac{MR^2}{2}\times \frac{a}{R}=\frac{MRa}{2}------ 1

Consider the free body diagram of the cylinder on the incline. The forces acting along the incline are mg\sin \theta\ and\ f. The net force acting along the incline is given as:

F_{net}=Mg\sin \theta-f\\But,\ f=\mu_s N\\So, F_{net}=Mg\sin \theta -\mu_s N-------- 2

Now, consider the forces acting perpendicular to the incline. As there is no motion in the perpendicular direction, net force is zero.

So, N=Mg\cos \theta

Plugging in N=Mg\cos \theta in equation (2), we get

F_{net}=Mg\sin \theta -\mu_s Mg\cos \theta\\F_{net}=Mg(\sin \theta-\mu_s \cos \theta)--------------3

Now, as per Newton's second law,

F_{net}=Ma\\Mg(\sin \theta-\mu_s \cos \theta)=Ma\\\therefore a=g(\sin \theta-\mu_s \cos \theta)------4

Now, torque acting on the cylinder is provided by the frictional force and is given as the product of frictional force and radius of the cylinder.

\tau=fR\\\frac{MRa}{2}=\mu_sMg\cos \theta\times  R\\\\a=2\times \mu_sg\cos \theta\\\\But, a=g(\sin \theta-\mu_s \cos \theta)\\\\\therefore g(\sin \theta-\mu_s \cos \theta)=2\times \mu_sg\cos \theta\\\\\sin \theta-\mu_s \cos \theta=2\mu_s\cos \theta\\\\\sin \theta=2\mu_s\cos \theta+\mu_s\cos \theta\\\\\sin \theta=3\mu_s \cos \theta\\\\\mu_s=\frac{\sin \theta}{3\cos \theta}\\\\\mu_s=\frac{1}{3}\tan \theta............(\because \frac{\sin \theta}{\cos \theta}=\tan \theta)

Therefore, the minimum coefficient of static friction needed for the cylinder to roll down without slipping is given as:

\mu_s=\frac{1}{3}\tan \theta

3 0
3 years ago
Read 2 more answers
A straight wire of length 0.62 m carries a conventional current of 0.7 amperes. What is the magnitude of the magnetic field made
anyanavicka [17]

Answer:

Magnetic field at point having a distance of 2 cm from wire is 6.99 x 10⁻⁶ T

Explanation:

Magnetic field due to finite straight wire at a point perpendicular to the wire is given by the relation :

B=\frac{\mu_{0}I }{2\pi R }\times\frac{L}{\sqrt{L^{2}+R^{2}  } }      ......(1)

Here I is current in the wire, L is the length of the wire, R is the distance of the point from the wire and μ₀ is vacuum permeability constant.

In this problem,

Current, I = 0.7 A

Length of wire, L = 0.62 m

Distance of point from wire, R = 2 cm = 2 x 10⁻² m = 0.02 m

Vacuum permeability, μ₀ = 4π x 10⁻⁷ H/m

Substitute these values in equation (1).

B=\frac{4\pi\times10^{-7}\times  0.7 }{2\pi \times0.02 }\times\frac{0.62}{\sqrt{(0.62)^{2}+(0.02) ^{2}  } }

B = 6.99 x 10⁻⁶ T

3 0
3 years ago
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