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DiKsa [7]
3 years ago
12

a tire is rolling along a road, without slipping with a velocity v. a piece of tape is attached to the tire. When the tape is op

posite the road (at the top of the tire), its velocity with respect to the road is
Physics
1 answer:
Kryger [21]3 years ago
4 0

Answer:

The right solution will be the "2v".

Explanation:

For something like an object underneath pure rolling the speed at any point is calculated by:

⇒  v_{rolling}=v_{translational}+v_{rotational}

Although the angular velocity was indeed closely linked to either the transnational velocity throughout particular instance of pure rolling as:

⇒  \omega=\frac{v_{translational}}{r}

Significant meaning is obtained, as speeds are in the same direction. Therefore the speed of rotation becomes supplied by:

⇒  v_{rotational}=\omega \times r

On substituting the estimated values, we get

⇒                   =\frac{v_{translational}}{r} \times r

⇒                   =v_{translational}

So that the velocity will be:

⇒  v_{rolling}=v+v

⇒              =2v

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A 55-kg block, starting from rest, is pushed a distance of 5.0 m across a floor by a horizontal force Fp whose magnitude is 140
max2010maxim [7]

Answer:

The answer is "151.25 J and -547.64 J".

Explanation:

u = 0\\\\v = 2.35\  \frac{m}{sec}\\\\d = 5.0 \ m\\\\

Using formula:

v^2 = u^2 + 2 \times a \times d\\\\2.35^2 = 0^2 + 2 \times a \times 5\\\\a = \frac{2.35^2}{10} \\\\

   = 0.55 \ \frac{m}{sec^2}\\\\

F_{net} = m \times a\\\\F_{net} = 55 \times 0.55 = 30.25\ N\\\\

Calculating the Work by net force

W = F_{net}\times d\\\\W = 30.25 \times 5 = 151.25 \ J\\\\

The above work is converted into thermal energy.

Now,

F_{net} = F_p - F_f\\\\F_p = 140 \ N\\\\F_f = u_k\times m \times g = u_k \times 55 \times 9.81\\\\F_f = 539.55 \times u_k\\\\30.25 = 140 - u_k \times 55 \times 9.81\\\\u_k = \frac{(140 - 30.25)}{(55\times 9.81)}\\\\uk = 0.203 = \text{Coefficient of friction}\\\\W_f = -F_f \times d\\\\W_f = -0.203 \times 55 \times 9.81 \times 5\\\\Work\ done\  by\ friction = -547.64 \ J

6 0
3 years ago
According to the theory of plate tectonics
Darya [45]

Answer:

Alfred Wegener proposed the idea that earths lithospher is divided into tectonic plates.The theory was only accepted after he died.

8 0
3 years ago
By how much does the pressure of a gas in a rigid vessel decrease when the temperature drops from 0 ∘C to –1∘C?
Ugo [173]
When a problem says a rigid vessel, it means that volume is constant. At constant V, pressure and temperature are indirectly proportional. We calculate as follows:

P1/T1 = P2/T2
P1/P2 = T1/T2
P1/P2 = 273.15 / 272.15
P1/P2 = 1.00

Hope this helps. Have a nice day.
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3 years ago
The parallax method of measuring star distances gives most accurate results when the gap between two observations of a star is a
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<span>b. It ensures that measurements are taken from two points
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Measurements taken six months apart are the farthest apart
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8 0
3 years ago
Read 2 more answers
An ideal monatomic gas initially has a temperature of 300 K and a pressure of 5.79 atm. It is to expand from volume 420 cm3 to v
maxonik [38]

Answer:

a) The final pressure is 1.68 atm.

b) The work done by the gas is 305.3 J.

Explanation:

a) The final pressure of an isothermal expansion is given by:

T = \frac{PV}{nR}

T_{i} = T_{f}

\frac{P_{i}V_{i}}{nR} = \frac{P_{f}V_{f}}{nR}

Where:

P_{i}: is the initial pressure = 5.79 atm

P_{f}: is the final pressure =?

V_{i}: is the initial volume = 420 cm³

V_{f}: is the final volume = 1450 cm³

n: is the number of moles of the gas

R: is the gas constant

P_{f} = \frac{P_{i}V_{i}}{V_{f}} = \frac{5.79 atm*420 cm^{3}}{1450 cm^{3}} = 1.68 atm

Hence, the final pressure is 1.68 atm.

b) The work done by the isothermal expansion is:

W = P_{i}V_{i}ln(\frac{V_{f}}{V_{i}}) = 5.79 atm*\frac{101325 Pa}{1 atm}*420 cm^{3}*\frac{1 m^{3}}{(100 cm)^{3}}ln(\frac{1450 cm^{3}}{420 cm^{3}}) = 305.3 J

Therefore, the work done by the gas is 305.3 J.

I hope it helps you!        

3 0
3 years ago
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