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3 years ago

1. As you get ready for bed, you roll up one of your socks into a tight ball and toss it into the laundry

Physics

1 answer:

yKpoI14uk [10]3 years ago

3 0

Answer:

When you toss a rolled up sock across the room, it travels faster as it becomes round and has more weight added on it and this causes the sock to travel in the direction you wish and this gives you a high chance of the sock going straight into the laundry basket, no matter how far away you are.

On the other hand, throwing a sock without rolling it up will cause the sock to just flat down as you throw it. It will travel at a low speed because it has no weight on it since it is flat, and if you try to throw it, it will atleast land 21 cm away from you. About a 0% chance of it getting in the basket.

Hope this helped! =>

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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 magnetic field is uniform over a flat, horizontal circular region with a radius of 1.50 mm, and the field varies with time. In

Daniel [21]

Answer: $81.57\mu V$

Explanation:

Given

radius of circular region r=1.50 mm

$A=\pi r^2=7.069\times 10^{-6}\ m^2$

Magnetic Field $B=1.50\ T$

time t=130 ms

Flux is given by

$\phi =B\cdot A$

change in Flux $d\phi =(B_f-B_i)A$

Emf induced is $e=\frac{\mathrm{d} \phi}{\mathrm{d} t}$

$e=\frac{(1.5)\cdot 7.069\times 10^{-6}}{130\times 10^{-3}}$

$e=81.57 \mu V$

3 0

3 years ago

A bungee jumper starts with 1000 J in their GPE store. After they jump they fall and are brought to a stop with the bungee cord.

pochemuha

Answer:

energy is equal to 1000 J

Explanation:

When the jumper is in the tent, he has a given height, this height gives him a gravitational potential energy, which forms his initial mechanical energy of 1000 J. After jumping, this energy is converted into elastic energy of the rope plus a remainder of potential energy gravitational, it does not reach the ground, but as the friction is negligible the total mechanical energy is conserved, therefore its energy is equal to 1000 J

This is a case of energy transformation, but the total value of mechanical energy does not change

8 0

3 years ago

A friend asks you how much pressure is in your car tires. You know that the tire manufacturer recommends 30 psi, but it's been a

Rama09 [41]

Answer:

21 psi

Explanation:

The weight of the car is:

W = mg

W = 1000 kg * 9.8 m/s²

W = 9800 N

Divided by 4 tires, each tire supports:

F = W/4

F = 9800 N / 4

F = 2450 N

Pressure is force divided by area, so:

P = F / A

P = (2450 N) / (0.13 m × 0.13 m)

P ≈ 145,000 Pa

101,325 Pa is the same as 14.7 psi, so:

P ≈ 145,000 Pa × (14.7 psi / 101,325 Pa)

P ≈ 21 psi

3 0

3 years ago

A capacitor is formed from two concentric spherical conducting shells separated by vacuum. The inner sphere has radius 12.0 cm ,

iren [92.7K]

Energy Density = 1/2 × ε(0) × (V/d)^2

V = 100, d = 0.01, ε(0) = 8.85 x 10^-12

5 0

2 years ago