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3 years ago

13

Please help me! It states constant speed but moving in circular motion. I know there is acceleration due to a rate of change of

velocity but what is the direction

1 answer:

Morgarella [4.7K]3 years ago

8 0

Direction of resultant force in circular motion is directed towards the center of the circle. Hence vector B is the direction of the resultant force.

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A capacitor with an initial potential difference of 100 V isdischarged through a resistor when a switch between them is closed a

jonny [76]

Answer:

(a). The time constant of the circuit is 2.17.

(b). The potential difference across the capacitor at t=17.0 s is 0.0396 V.

Explanation:

Given that,

Initial potential difference = 100 V

Potential difference across the capacitor = 1.00 V

(a). We need to calculate the time constant of the circuit

Using formula of potential difference

$V(t)=V_{0}e^{\dfrac{-t}{RC}}$

Put the value into the formula

$1.00=100e^{\dfrac{-10.0}{RC}}$

$0.01=e^{\dfrac{-10.0}{RC}}$

On taking ln

$ln(0.01)=\dfrac{-10}{RC}$

$RC=\dfrac{-10}{ln(0.01)}$

$RC=2.17$

(b). We need to calculate the potential difference across the capacitor at t=17.0 s

Using formula again

$V(17)=100e^{\dfrac{-17}{2.17}}$

$V{17}=0.0396\ V$

Hence, (a). The time constant of the circuit is 2.17.

(b). The potential difference across the capacitor at t=17.0 s is 0.0396 V.

7 0

3 years ago

HELP ME ASAP PLEASE

padilas [110]

Answer:

( 1000 × 4 = 4,000) (800×3= 2400) (800×2=1600) the answer is 1600 hope it helps

6 0

3 years ago

A boy reaches out of a window and tosses a ball straight up with a speed of 10 m/s. The ball is 20 m above the ground as he rele

kirill115 [55]

Explanation:

It is given that,

Speed of the ball, v = 10 m/s

Initial position of ball above ground, h = 20 m

(a) Let H is the maximum height reached by the ball. It can be calculated using the conservation of energy as :

$\dfrac{1}{2}mv^2=mgh'$

$h'=\dfrac{v^2}{2g}$

$h'=\dfrac{(10)^2}{2\times 9.8}$

h' = 5.1 m

The maximum height above ground,

H = 5.1 + 20

H = 25.1 meters

So, the maximum height reached by the ball is 25.1 meters.

(b) The ball's speed as it passes the window on its way down is same as the initial speed i.e. 10 m/s.

Hence, this is the required solution.

6 0

3 years ago

A man of weight Wman is standing on the second floor and is pulling on a rope to lift a box of weight Wbox from the floor below.

slavikrds [6]

Answer:

Explanation:

See the attached figure . See the forces acting on man pulling up the box .

Man is stationary so net force acting on man is zero .

T + R = Wman

R is the reaction force of the ground of second floor .

R = Wman - T

3 0

3 years ago

Tennis balls traveling at greater than 100 mph routinely bounce off tennis rackets. At some sufficiently high speed, however, th

Kipish [7]

Answer:

Probability of tunneling is $10^{- 1.17\times 10^{32}}$

Solution:

As per the question:

Velocity of the tennis ball, v = 120 mph = 54 m/s

Mass of the tennis ball, m = 100 g = 0.1 kg

Thickness of the tennis ball, t = 2.0 mm = $2.0\times 10^{- 3}\ m$

Max velocity of the tennis ball, $v_{m} = 200\ mph$ = 89 m/s

Now,

The maximum kinetic energy of the tennis ball is given by:

$KE = \frac{1}{2}mv_{m}^{2} = \frac{1}{2}\times 0.1\times 89^{2} = 396.05\ J$

Kinetic energy of the tennis ball, KE' = $\frac{1}{2}mv^{2} = 0.5\times 0.1\times 54^{2} = 154.8\ m/s$

Now, the distance the ball can penetrate to is given by:

$\eta = \frac{\bar{h}}{\sqrt{2m(KE - KE')}}$

$\bar{h} = \frac{h}{2\pi} = \frac{6.626\times 10^{- 34}}{2\pi} = 1.0545\times 10^{- 34}\ Js$

Thus

$\eta = \frac{1.0545\times 10^{- 34}}{\sqrt{2\times 0.1(396.05 - 154.8)}}$

$\eta = \frac{1.0545\times 10^{- 34}}{\sqrt{2\times 0.1(396.05 - 154.8)}}$

$\eta = 1.52\times 10^{-35}\ m$

Now,

We can calculate the tunneling probability as:

$P(t) = e^{\frac{- 2t}{\eta}}$

$P(t) = e^{\frac{- 2\times 2.0\times 10^{- 3}}{1.52\times 10^{-35}}} = e^{-2.63\times 10^{32}}$

$P(t) = e^{-2.63\times 10^{32}}$

Taking log on both the sides:

$logP(t) = -2.63\times 10^{32} loge$

$P(t) = 10^{- 1.17\times 10^{32}}$

6 0

3 years ago