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2 years ago

A 2.0kg mass is intally moving at a speed of 10m/s due north, when a constant

Physics

1 answer:

Gnom [1K]2 years ago

8 0

Answer:

Option(B) 56. east of north

Explanation:

2D Motion

If objects are moving in an XY plane subject to net force (F), acceleration (a), velocity (v) and displacement (r) in both axes, we must consider all those magnitudes as vectors because they have components in x and y.

If $\vec a$ is the acceleration vector, then

$\vec v=\vec v_o+\vec a.t$

$\vec r=\vec v_o.t+\frac{\vec a.t^2}{2}+\vec r_o$

$\vec F=m.\vec a$

Assuming the positive direction to the right and upwards, we are given the following data

$m=2\ kg$

$\vec v_o=0\ m/s\ \hat i+10\ \hat j$

The x-component of the velocity is zero because it due north.

The force is applied eastward;

$\vec F=10\ \hat i+0\ N\ \hat j$

t=3 sec

From the formula

$\vec F=m.\vec a$

we can solve for $\vec a$

$\displaystyle \vec a=\frac{\vec F}{m}$

$\displaystyle \vec a=\frac{10\ \hat i+0\ N\ \hat j}{2\ kg}$

$\displaystyle \vec a=5\ m/sec^2\ \hat i+0\ \hat j$

We can now compute the velocity at t=3 sec

$\vec v=0\ \hat i+10\ m/s\ \hat j+\ (5\ m/sec^2\ \hat i+0\ \hat j)3\ sec$

Adding and simplifying

$\vec v=15\ m/s\ \hat i+10\ m/s\ \ \hat j$

The direction is given by the angle computed as

$\displaystyle tan\theta=\frac{v_y}{v_x}$

$\displaystyle tan\theta=\frac{10}{15}$

$\theta=arctan(0.667)$

$\theta=33.7^o\approx 34^o$

This angle is north of east, the required angle is

$90^o-34^o=56^o$

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If displacement covered by a particle is zero then distance cover by it

sleet_krkn [62]

Answer:

When displacement is zero, the particle may be at rest, therefore, distance travelled = 0.

Again, when displacement is zero, the final position matches with the initial position after some time, but the distance travelled will not be zero.

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2 years ago

A light ray incident from medium 1 to medium 2, where n1>n2. When the incident angle exceed the critical angle ac, the refrac

vovikov84 [41]

Explanation:

(a)

Critical angle is the angle at the angle of refraction is 90°. After the critical angle, no refraction takes place.

Using Snell's law as:

$n_1\times {sin\theta_i}={n_2}\times{sin\theta_r}$

Where,

${\theta_i}$ is the angle of incidence

${\theta_r}$ is the angle of refraction = 90°

${n_2}$ is the refractive index of the refraction medium

${n_1}$ is the refractive index of the incidence medium

Thus,

$n_1\times {sin\ \theta_{critical}}={n_2}\times{sin\ 90^0}$

The formula for the calculation of critical angle is:

${sin\theta_{critical}}=\frac {n_2}{n_1}$

Where,

${\theta_{critical}}$ is the critical angle

(b)

No it cannot occur. It only occur when the light ray bends away from the normal which means that when it travels from denser to rarer medium.

7 0

3 years ago

A 6- F capacitor is charged to 90 V and is then connected across a 700- resistor. What is the initial charge on the capacitor

Lorico [155]

Answer:

540C.

Explanation:

A capacitor of capacitance C when charged to a voltage of V will have a charge Q given as follows;

Q = CV ----------(i)

From the question, the initial charge on the capacitor is the charge on it before it was connected to the resistor. In other words, the initial charge on the capacitor will have a maximum value which can be calculated using equation (i) above.

Where;

C = 6F

V = 90V

Substitute these values into equation (i) as follows;

Q = 6 x 90

Q = 540 C

Therefore, the initial charge on the capacitor is 540C.

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2 years ago

A fan at a rock concert is 50.0 m from the stage, and at this point the sound intensity level is 114 dB. Sound is detected when

Marianna [84]

Answer:

A) $P=13.92\ J.s^{-1}$

B) $v=3730.9912\ m.s^{-1}$

C) $v=74.44\ mm.s^{-1}$

D) mosquitoes speed in part B is very much larger than that of part C.

Explanation:

Given:

Distance form the sound source, $s=50\ m$

sound intensity level at the given location, $\beta=114\ dB$

diameter of the eardrum membrane in humans, $d=8.4 \times 10^{-3}\ m$

We have the minimum detectable intensity to the human ears, $I_0=10^{-12}\ W.m^{-2}$

(A)

Now the intensity of the sound at the given location is related mathematically as:

$\beta=10\ log(\frac{I}{I_0} )$ ..........................................(1)

$114=10\ log\ (\frac{I}{10^{-12}} )$

$11.4=log\ I+12\ log\ 10$

$I=0.2512\ W.m^{-2}$

As we know :

$I=\frac{P}{A}$

$0.2512=\frac{P}{\pi\times \frac{8.4^2}{4} }$

$P=13.92\ J.s^{-1}$ is the energy transferred to the eardrums per second.

(B)

mass of mosquito, $m=2\times 10^{-6}\ kg$

Now the velocity of mosquito for the same kinetic energy:

$KE=\frac{1}{2} m.v^2$

$13.92=\frac{1}{2}\times 2\times 10^{-6}\times v^2$

$v=3730.9912\ m.s^{-1}$

(C)

Given:

Sound intensity, $\beta = 20\ dB$

Using eq. (1)

$20=10\ log\ (\frac{I}{10^{-12}} )$

$2=log\ I+12\ log\ 10$

$I=10^{-10}\ W.m^{-2}$

Now, power:

$P=I.A$

$P=10^{-10}\times \pi\times \frac{8.4^2}{4}$

$P=5.54\times 10^{-9}\ J.s^{-1}$

Hence:

$KE=\frac{1}{2} m.v^2$

$5.54\times 10^{-9}=0.5\times 2\times 10^{-6}\times v^2$

$v=0.07444\ m.s^{-1}$

$v=74.44\ mm.s^{-1}$

(D)

mosquitoes speed in part B is very much larger than that of part C.

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3 years ago

a truck was traveling at 16.6 miles per second and accelerates at a rate of 2.0 meters per second squared how much time is requi

Ivan

A truck was traveling at 16.6 miles per second and accelerates at a rate of 2.0 meters per second squared then time is required for the truck to reach a speed of 25 miles per second is 6759 s.

Explanation:

Velocity is defined as the rate of change in displacement while acceleration is defined as the rate of change of velocity. Acceleration may be positive or negative. Acceleration is positive when the velocity of the object is increases and it is negative when velocity of the object is decreases. Negative acceleration is also called deceleration.

Mathematically

a = $\frac{(v_{f} - v_{i})}{t}$

Where a is the acceleration of the object, $v_{f}$ is the final velocity of the object and $v_{i}$ is the initial velocity of the object. t is equal to time taken.

Given data:

$v_{f}$ = 25 miles/s

$v_{i}$ = 16.6 miles/s

a = 2.0 m/s²

t = ?

As velocities and acceleration given in different units, So we need to convert to obtain same units. Here we convert unit of acceleration from m/s² to miles/s².

1 m/s² = 0.000621371192 miles/s²

2 m/s² = 0.00124274238 miles/s²

So,

a = 0.00124274238 miles/s²

Apply formula

a = $\frac{v_{f} - v_{i}}{t}$