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3 years ago

How much time is required for a car to reach a speed of 100 m/s if it starts from rest and its average acceleration is 12 m/s2?

Physics

2 answers:

gtnhenbr [62]3 years ago

7 0

Answer:

t = 8.3s

Explanation:

V = 100 m/s U = 0 m/s t = ? a = 12 m/s2

V = U + at

100 = 0 + 12 × t = 100 = 12t

12t = 100

t = 100/12 = 25/3

t = 8.3s

harkovskaia [24]3 years ago

5 0

A :-) for this question , we should apply
a = v - u by t
t = v - u by a
Given - u = 0
v = 100 m/s
a = 12 m/s^2
Solution -
t = v - u by a
t = 100 - 0 by 12
t = 100 by 12
( cut 12 and 100 because 2 x 50 = 100 ,
2 x 6 = 12 )
( cut 6 and 50 because 2 x 3 = 6 ,
2 x 25 = 50 )
t = 3 by 25
t = 0.12 sec

.:. The time is 0.12 sec.

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What is the formula for displacement

pogonyaev

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3 years ago

A typical incandescent light bulb consumes 75 W of power and has a mass of 30 g. You want to save electrical energy by dropping

nadya68 [22]

Answer:

The height will be 917431.2 m.

Explanation:

Power of bulb = 75 W

Time kept on 1 hr = 60 x 60 = 3600 sec

Energy of bulb = power x time

E = 75 x 3600 = 270000 J

From conservation of energy, kinetic energy of the bulb is equal to the potential energy of the bulb due to its height of fall.

Potential energy = m x g x h

Where g = acceleration due to gravity 9.81 m/s2

m = mass = 30 g = 0.03 kg

PE = 0.03 x 9.81 x h = 0.2943h

Equating withe the energy of bulb (still obeying energy conservation)

270000 = 0.2943h

h = 270000/0.2943 = 917431.2 m

8 0

4 years ago

Problem: The circular blad on a radial arm saw is turning at262

kobusy [5.1K]

Answer:

Net torque, $\tau=-0.033\ N-m$

Explanation:

It is given that,

Initial angular speed of the blade, $\omega_i=262\ rad/s$

Final angular speed of the blade, $\omega_f=85\ rad/s$

Time, t = 18 s

Radius of the disk, r = 0.13 m

Mass of the disk, m = 0.4 kg

We need to find the net torque applied to the blade. We know that in rotational mechanics the net torque acting on an object is equal to the product of moment of inertia and the angular acceleration such that,

$\tau=I\times \alpha$

The moment of inertia of the disk, $I=\dfrac{mr^2}{2}$

$\tau=\dfrac{mr^2}{2}\times \dfrac{\omega_f-\omega_i}{t}$

$\tau=\dfrac{0.4\times (0.13)^2}{2}\times \dfrac{85-262}{18}$

$\tau=-0.033\ N-m$

Negative sign shows that the net torque is acting in the opposite direction of its motion. Hence, this is the required solution.

3 0

3 years ago

Two blocks of weight 3.0N and 7.0N are connected by a massless string and slide down a 30 degree inclined plane. The coefficient

Luba_88 [7]

Answer:

a. a = $6.41 m/s^2$

b. T = -0.81 N

Explanation:

Given,

weight of the lighter block = $w_1\ =\ 3.0\ N$
weight of the heavier block = $w_2\ =\ 7.0\ N$
inclination angle = $\theta\ =\ 30^o$
coefficient of kinetic friction between the lighter block and the surface = $\mu_1\ =\ 0.13$

coefficient of kinetic friction between the heavier block and the surface = $\mu_2\ =\ 0.31$
friction force on the lighter block = $f_1\ =\ \mu_1N_1\ =\ \mu_1 w_1cos\theta$
friction force on the heavier block = $f_2\ =\ \mu_2N_2\ =\ \mu_2w_2cos\theta$

Let 'a' be the acceleration of the blocks and 'T' be the tension in the string.

From the f.b.d. of the lighter block,

$w_1sin\theta\ -\ T\ -\ f_1\ =\ \dfrac{w_1a}{g}\\\Rightarrow T\ =\ w_1sin\theta\ -\ \dfrac{w_1a}{g}\ -\ f_1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,eqn (1)$

From the f.b.d. of the heavier block,

$w_2sin\theta\ +\ T\ -\ f_2\ =\ \dfrac{w_2a}{g}\\\Rightarrow T\ =\ \dfrac{w_2a}{g}\ -\ w_2sin\theta\ -\ f_2\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,eqn (2)$

From eqn (1) and (2), we get,

$w_1sin\theta\ -\ \dfrac{w_1a}{g}\ -\ f_1\ =\ \dfrac{w_2a}{g}\ -\ w_2sin\theta\ -\ f_2\\\Rightarrow w_1gsin\theta \ -\ w_1a\ -\ \mu_1w_1gcos\theta\ =\ w_2a\ -\ w_2gsin\theta\ -\ \mu_2 w_2gcos\theta\\$

$\Rightarrow a\ =\ \dfrac{g(w_1sin\theta\ -\ \mu_1w_1cos\theta\ +\ w_2sin\theta\ +\ \mu_2w_2cos\theta)}{w_1\ +\ w_2}\\\Rightarrow a\ =\ \dfrac{g[sin\theta(w_1\ +\ w_2)\ +\ cos\theta(\mu_2w_2\ -\ \mu_1w_1)]}{w_1\ +\ w_2}\\$

$\Rightarrow a\ =\ \dfrac{9.81\times [sin30^o\times (3.0\ +\ 7.0)\ +\ cos30^o\times (0.31\times 7.0\ -\ 0.13\times 3.0)]}{3.0\ +\ 7.0}\\\Rightarrow a\ =\ 6.4\ m/s.$

part (b)

From the eqn (2), we get, $T\ =\ \dfrac{w_2a}{g}\ -\ w_2sin\theta\ -\ \mu_2w_2cos\theta\\\Rightarrow T\ =\ \dfrac{7.0\times 6.41}{9.81}\ -\ 7.0\times sin30^o\ -\ 0.31\times 7.0\times cos30^o\\\Rightarrow T\ =\ -0.81\ N$

3 0

3 years ago

Read 2 more answers

A student boils water in a can and puts the hot can carefully in very icy water. The can immediately crushes due to the change i

hammer [34]

Answer:

By elimination I know it's not CGL and I know it's not Charle's Law.

Gay-Lussac's law states the pressure of a a given mass of gas varies directly with the absolute temperature of the gas, when the volume is kept constant. I think it's Gay-Lussac's Law.

Boyle's law states that pressure of a gas tends to increase as the volume of the container decreases which I got from a google search because I didn't know what it was.

CGL is the combined formula of all of these laws by the way

Charle's law is just gas expands when heated.

Gay-Lussac's is the best answer choice in my opinion

8 0

3 years ago