You attach a meter stick to an oak tree, such that the top of the meter stick is 1.87 meters above the ground. Later, an acorn falls from somewhere higher up in the tree. If the acorn takes 0.166 seconds to pass the length of the meter stick, how high above the ground was the acorn before it fell (assuming that the acorn didn\'t run into any branches or leaves on the way down)?

Answer:

Electric field at distance of 50 micrometer due to a proton is 0.576 N/C

Explanation:

We have given distance where we have to find the electric field [tex]r=50\mu m=50\times 10^{-6}m[/tex]

We know that charge on proton [tex]q=1.6\times 10^{-16}C[/tex]

Electric field due to point charge is given by [tex]E=\frac{Kq}{r^2}[/tex], here K is a constant which value is [tex]9\times 10^9Nm^2/C^2[/tex]

So electric field [tex]E=\frac{9\times 10^9\times 1.6\times 10^{-16}}{(50\times 10^{-6})^2}=0.576N/C[/tex]

Answer:

All of the above

Explanation:

The correct answer is option E (All of the above)

All the options are alternative source of energy.

The option given are not the traditional way (energy production from coal) of extracting energy as the loss of natural resources does not take place in these source of energy.

energy extracted from wind,  solar light , hydrogen ,etc are the source of energy  the alternative source of production of energy because do not exploit the natural resources, reduce the carbon emission and energy produced by them is clean energy.

Answer:

[tex]F_{net}=N\ sin\theta[/tex]

Explanation:

Let a car of m is on an incline bank of angle θ and it is rounding a curve with no friction. We need to find the centripetal force acting on it.    

The attached free body diagram shows the car on the banked turn. It is clear that,

In vertical direction,

[tex]N\ cos\theta=mg[/tex]

In horizontal direction,

[tex]F_{net}=F_{centripetal}[/tex]

[tex]F_{net}=N\ sin\theta[/tex]

So, the centripetal force is equal to [tex]N\ sin\theta[/tex]. Hence, the correct option is (c).

Answer:

The angle is [tex]\theta\approx 14.61[/tex] degrees.

Explanation:

Se the attached drawing if you need a visual aid for the explanation. Let [tex]\theta[/tex] be the angle of elevation of the plante which in itself is the same drop angle that the pilot measures. Let [tex]d[/tex] be the horizontal distance from the target and [tex]h[/tex] the height of the plane. We know that the package is dropped without any initial vertical speed, that means that it has a y-position equation of the form:

[tex]y(t)=-\frac{1}{2}gt^2+h[/tex]

If we set [tex]y(t)=0[/tex] we are setting the condition that the package is in the ground. We can then solve for t and get the flight time of the package.

[tex]0=-\frac{1}{2}gt^2+h\implies t_f=\sqrt{\frac{2h}{g}}[/tex].

If the flight time is -[tex]t_f[/tex] then the distance b can be found in meters by taking into account that the horizontal speed of the plane is [tex]v=283\, Km/h=78.61 \, m/s[/tex].

[tex]d=v\cdot t_f=78.61\cdot \sqrt{\frac{2h}{g}}[/tex]

The angle is thus

[tex]\theta=\arctan{\frac{h}{v\cdot t_f}}=\arctan{\frac{h}{v\cdot \sqrt{\frac{2\cdot h}{g}}}\approx 14.61 [/tex] degrees.

Answer:48 %

Explanation:

Given

outside temperature [tex]=25^{\circ}\approx 298 K[/tex]

Internal temperature[tex]=300^{\circ}\approx 573 K[/tex]

Maximum efficiency is carnot efficiency which is given by

[tex]\eta =1-\frac{T_L}{T_H}[/tex]

[tex]\eta =1-\frac{298}{573}[/tex]

[tex]\eta =\frac{275}{573}=0.4799[/tex]

i.e. [tex]47.99 %\approx 48 %[/tex]

No it is not possible to achieve this efficiency because carnot engine is the ideal engine so practically it is not possible but theoretically it is possible

Answer:

967500000 years

Explanation:

The Speed at which the radius of the orbit of the Moon is increasing is 4 cm/yr

Converting to m

1 m = 100 cm

[tex]1\ cm=\frac{1}{100}\ m[/tex]

[tex]4\ cm\y=\frac{4}{100}=0.04\ m/yr[/tex]

The distance by which the radius increases is 3.84×10⁷ m

Time = Distance / Speed

[tex]\text{Time}=\frac{3.87\times 10^7}{0.04}\\\Rightarrow \text{Time}=967500000\ yr[/tex]

967500000 years will pass before the radius of the orbit increases by 10%.

Final answer:

The radius of the Moon's orbit increases by approximately 4 cm/year. To calculate the number of years it will take for the radius to increase by 3.84 × 10^6 m, we can use the formula Time = Change in Distance / Rate of Increase. The answer is approximately 9.6 × 10^7 years.

Explanation:

The radius of the Moon's orbit is increasing at a rate of approximately 4 cm/year. To find out how many years will pass before the radius of the Moon's orbit increases by 3.84 × 10^6 m, we can use the formula:
Time = Change in Distance / Rate of Increase
Substituting the given values, we get:
Time = (3.84 × 10^6 m) / (4 cm/year)
Now, we need to convert the units so that they are consistent. 3.84 × 10^6 m is equivalent to 3.84 × 10^8 cm. Substituting this value into the equation, we get:
Time = (3.84 × 10^8 cm) / (4 cm/year)
Canceling out the units, we find that:
Time = 9.6 × 10^7 years.

Answer:

The angular frequency [tex]\omega[/tex] of the oscillation is [tex]0.58s^{-1}[/tex]

Explanation:

For this particular situation, the angular frequency of the system is given by

[tex]\omega=\sqrt{\frac{m_1+m_2}{m_1m_2}k}=\sqrt{\frac{7 kg+5 kg }{7kg *5 kg}1\frac{N}{m}}=\sqrt{\frac{3}{35s^2}}\approx 0.58s^{-1}[/tex]

Answer: 208.3 s

Explanation:

Hi!

You need to calculate the velocity of the boat relative to the shore, in each trip.

Relative velocities are transformed according to:

[tex]V_{b,s} = V_{b,w} + V_{w,s}\\ V_{b,s} = \text{velocity of boat relative to shore}\\V_{b,w} = \text{velocity of boat relative to water} \\V_{b,w} = \text{velocity of water relative to shore}\\[/tex]

Let's take the upstram direction as positive. Water flows downstream, so it's velocity relative to shore is negative , -2 m/s

In the upstream trip, velocity of boat relative to water is positive: 10m/s. But in the downstream trip it is negatoive: -10m/s

In upstream trip we have:

[tex]V_{b,s} = (10 - 2)\frac{m}{s} = 8 ms[/tex]

In dowstream we have:

[tex]V_{b,s} = (-10 - 2)\frac{m}{s} = -12 ms[/tex]

In both cases the distance travelled is 1000m. Then the time it takes the round trip is:

[tex]T = T_{up} + T_{down} = \frac{1000}{8}s + \frac{1000}{12}s = 208.3 s[/tex]

Firstly , let's consider the motion of the ball as it travelled up past the window. Given that the distance covered by the ball is 1.19 m in 0.20 s, we can calculate the initial velocity of the ball using the straightforward kinematic equation for velocity, which is v = d/t. Here, 'd' is the distance travelled by the ball, and 't' is the time taken to cover that distance.

Plugging the values into the equation:
 vi = d/t
 vi = 1.19 m / 0.20 s
 vi = 5.95 m/s

Therefore, the initial velocity of the ball is approximately 5.95 m/s.

Next, we want to determine how long it would take for the ball to reach its peak and then reappear. The acceleration due to gravity (g) is 9.81 m/s^2 and acts downwards. Meaning that, as the ball ascends, it slows down until it reaches its peak, where its velocity becomes zero.

To find the time at peak, we can use the equation of motion v = vi + at, where 'v' is the final velocity, 'vi' is the initial velocity, 'a' is the acceleration, and 't' is time. At the peak, the final velocity 'v' becomes 0, thus:

t_peak = vi / g
t_peak = 5.95 m/s / 9.81 m/s²
t_peak = 0.61 s

Therefore, the time it takes to reach the peak is approximately 0.61 seconds.

As we're dealing with symmetrical motion (upwards and downwards motion are equal), the time for the ball to descend from its peak would be identical to the time it took to ascend, which means:

Time for the ball to reappear, t_reappear = 2 * t_peak
t_reappear = 2 * 0.61 s
t_reappear = 1.22 s

Hence, it should take approximately 1.22 seconds before the ball reappears.

https://brainly.com/question/29545516

#SPJ12

Explanation:

Given that,

Charge 1, [tex]q_1=7.8\ nC=7.8\times 10^{-9}\ C[/tex]

Charge 2, [tex]q_2=4.3\ nC=4.3\times 10^{-9}\ C[/tex]

Distance between charges, r = 1.81 m

(a) The electrostatic force that one charge exerts on the another is given by :

[tex]F=k\dfrac{q_1q_2}{r^2}[/tex]

[tex]F=9\times 10^9\times \dfrac{7.8\times 10^{-9}\times 4.3\times 10^{-9}}{(1.81)^2}[/tex]

[tex]F=9.21\times 10^{-8}\ N[/tex]

(b) As both charges are positively charged. So, the force of attraction between them is repulsive.

Answer:

a) 35.44 mm

b) 17.67 mm

Explanation:

u = Object distance =  3.6 m

v = Image distance

f = Focal length = 35 mm

[tex]h_u[/tex]= Object height = 1.8 m

a) Lens Equation

[tex]\frac{1}{f}=\frac{1}{u}+\frac{1}{v}\\\Rightarrow \frac{1}{f}-\frac{1}{u}=\frac{1}{v}\\\Rightarrow \frac{1}{v}=\frac{1}{35}-\frac{1}{3600}\\\Rightarrow \frac{1}{v}=\frac{713}{25200} \\\Rightarrow v=\frac{25200}{713}=35.34\ mm[/tex]

The CCD sensor is 35.34 mm from the lens

b) Magnification

[tex]m=-\frac{v}{u}\\\Rightarrow m=-\frac{35.34}{3600}[/tex]

[tex]m=\frac{h_v}{h_u}\\\Rightarrow -\frac{35.34}{3600}=\frac{h_v}{1800}\\\Rightarrow h_v=-\frac{35.34}{3600}\times 1800=-17.67\ mm[/tex]

The person appears 17.67 mm tall on the sensor

Answer:

V = 7.91 m/s(magnitude) @ 72.3°(direction)

Explanation:

We will have to do a parabolic movement analysis in order to determine the initial velocity. For this we will mainly use the following kinetic formulae:

[tex]1) X_{f}=X_{0}+V_{0}t+\frac{at^{2}}{2}\\\\2) V_{f}=V_{0}+at[/tex]

We will work backwards from the initial information: a final velocity of 11.5 m/s at a 78.2° angle. From this we can extract the final horizontal and vertical velocity:

[tex]V_{fx}=11.8\frac{m}{s}*Cos(78.2)=2.41\frac{m}{s}\\\\V_{fy}=11.8\frac{m}{s}*Sin(78.2)=11.55\frac{m}{s}[/tex]

We will use [tex]V_{fy}[/tex] with a negative value (since the diver is moving in a downwards direction). That way we can find the time elapsed between the highest point of the trayectory to instants before she contacts the water. At the maximum height of the movement, the velocity in the vertical direction is zero. Therefore we can rewrite equation 2) and solve for t:

[tex]-11.55\frac{m}{s}=0\frac{m}{s} - 9.8\frac{m}{s^{2}}t\\\\t= \frac{-11.55\frac{m}{s}}{- 9.8\frac{m}{s^{2}}}=1.17s[/tex]

This t that we found represents the time from the maximum height to the surface of the water. Now we can use equartion 1) to determine the maximum height. We must remember a starting point of 3.9m above the water:

[tex]0 = h+3.9m+0\frac{m}{s}*t-\frac{9.8\frac{m}{s^{2}}t^{2}}{2}\\ \\h=\frac{9.8\frac{m}{s^{2}}1.17s^{2}}{2}-3.9m=2.9m\\\\[/tex]

Since we now know what is the maximum height we can determine the time that it took for the diver to reach that point using equation 1) and 2):

[tex]2.9m =V_{0y}t-\frac{9.8\frac{m}{s^{2}}t^{2}}{2}\\\\0\frac{m}{s}= V_{0y}t-9.8\frac{m}{s^{2}}t[/tex]

We will solve for [tex]V_{ox}[/tex] in equation 2) and replace it in equation 1):

[tex]V_{0y} = 9.8\frac{m}{s^{2}}t\\\\2.9m =(9.8\frac{m}{s^{2}}t)t-\frac{9.8\frac{m}{s^{2}}t^{2}}{2}[/tex]

We solve for t (time from diving board to maximum height):

[tex]2.9m =(9.8\frac{m}{s^{2}}t)t-\frac{9.8\frac{m}{s^{2}}t^{2}}{2}\\\\2.9m=\frac{9.8\frac{m}{s^{2}}t^{2}}{2}\\\\5.8m=9.8\frac{m}{s^{2}}t^{2}\\\\t=\sqrt{ \frac{5.8m}{9.8\frac{m}{s^{2}}}}=0.77s[/tex]

Finally we can replace our t value in [tex]V_{0y}t=9.8\frac{m}{s^{2}}t[/tex] to obtain:

[tex]V_{0y}t=9.8\frac{m}{s^{2}}t=9.8\frac{m}{s^{2}}*0.77s=7.54\frac{m}{s}[/tex]

With this we have the information we need in order to determine the initial velocity of the diver. The magnitude will be calculated from the initial horizontal and vertical velocities (consider final and initial horizontal velocity to be equal in PARABOLIC MOVEMENTS):

[tex]V_{0}=\sqrt{V_{0x}^{2}+V_{0y}^{2}}=\sqrt{(2.41\frac{m}{s})^{2}+(7.54\frac{m}{s})^{2}}=7.91\frac{m}{s}[/tex]

And the direction is determined by an arctangent:

[tex]tan\theta=\frac{7.54\frac{m}{s}}{2.41\frac{m}{s}}\\\\\theta=arctan(\frac{7.54\frac{m}{s}}{2.41\frac{m}{s}})=72.3[/tex]

Final answer:

To find the diver's initial velocity, use equations involving vertical and horizontal components considering her final speed and displacement. The magnitude is 15.93 m/s, and the direction is horizontal.

Explanation:

To determine the diver's initial velocity, we first find the vertical component of her velocity using the equation vf=vi+at. Here, vf = 11.8 m/s, vi is the vertical component of the initial velocity, a = -9.81 m/s^2, and t = time in the air. Solving for vi, we get vi = 15.93 m/s. Then, to find the horizontal component of the initial velocity, we use the equation dx=vit+1/2at^2. Since the diver contacts the water at the highest point, her vertical displacement is 3.9m, and we can find the horizontal component of the initial velocity to be 0 m/s. Therefore, her initial velocity magnitude is 15.93 m/s, and her initial velocity direction is horizontal.

Answer:

(a) [tex]E_{ c} = 112.5 \mu J[/tex]

(b) [tex]E'_{ c} = 18 \mu J[/tex]

Solution:

According to the question:

Capacitance, C = [tex]1.00\mu F = 1.00\times 10^{- 6} F[/tex]

Voltage of the battery, [tex]V_{b} = 15.0 V[/tex]

(a)The Energy stored in the Capacitor is given by:

[tex]E_{c} = \frac{1}{2}CV_{b}^{2}[/tex]

[tex]E_{c} = \frac{1}{2}\times 1.00\times 10^{- 6}\times 15.0^{2}[/tex]

[tex]E_{ c} = 112.5 \mu J[/tex]

(b) When the voltage of the battery is 6.00 V, the the energy stored in the capacitor is given by:

[tex]E'_{c} = \frac{1}{2}CV'_{b}^{2}[/tex]

[tex]E'_{c} = \frac{1}{2}\times 1.00\times 10^{- 6}\times 6.0^{2}[/tex]

[tex]E'_{ c} = 18 \mu J[/tex]

(a) The energy stored in the capacitor is [tex]1.125 \times 10^{-4} \ J[/tex]

(b) The energy stored in the capacitor is [tex]1.8 \times 10^{-5} \ J[/tex]

The given parameters;

The energy stored in the capacitor is calculated as follows;

[tex]E = \frac{1}{2} CV^2\\\\E = \frac{1}{2} \times (1\times 10^{-6}) \times 15^2\\\\E = 1.125 \times 10^{-4} \ J[/tex]

When the battery voltage changes to 6 V, the energy stored in the capacitor is calculated as follows;

[tex]E = \frac{1}{2} CV^2\\\\E = \frac{1}{2} \times (1\times 10^{-6}) \times 6^2\\\\E = 1.8 \times 10^{-5} \ J\\\\[/tex]

Learn more here:https://brainly.com/question/24943210

Answer:

Statement E is false.

Explanation:

Statement a says that the object would have moved a distance of 10 m.

At t = 0

x0  = 0

y0 = 0

At t= 2

x2 = 2 * 2^2 = 8

y2 = 3 * 2 = 6

[tex]d2 = \sqrt{x2^2 + y2^2}[/tex]

[tex]d2 = \sqrt{8^2 + 6^2} = 10 m[/tex]

Statement A is true.

Statement B says that the momentum of the object would be 4*i+1.5*j

To test this we need to know the speed.

The speed is the first derivative by time:

v = 4*t*i + 3*j

At t = 2 the speed is

v2 = 8*i + 3*j

The momentum is

P = m * v

P2 = 0.5 * (8*i + 3*j) = (4*i+1.5*j)

This statement is true.

Statement C says that the kinetic energy of the object would be 18 J.

The magnitude of the speed is

[tex]v = \sqrt{8^2 + 3^2} = 8.54 m/s[/tex]

The kinetic energy is

Ek = 1/2 * m * v^2

Ek = 1/2 * 0.5 * 8.54^2 = 18 J

Statement C is true

Statement D says that the force acts perpenducular to the direction of the unit vector j.

The acceleration is the second derivative respect of time

a = 4*i + 0*j

Since there is no acceleration in the direction of j we can conclude that the force is perpendicular to the direction of j.

Statement D is true.

Statement E says that the force acting on the object is of 4 N

f = m * a

f = 0.5 * 4 = 2 N

Statement E is false.

Ans:

12500 N/C

Explanation:

Side of square,  a = 2.42 m

q = 4.25 x 10^-6 C

The formula for the electric field is given by

[tex]E = \frac{Kq}{r^2}[/tex]

where, K be the constant = 9 x 10^9 Nm^2/c^2 and r be the distance between the two charges

According to the diagram

BD = [tex]\sqrt{2}\times a[/tex]

where, a be the side of the square

So, Electric field at B due to charge at A

[tex]E_{A}=\frac{Kq}{a^2}[/tex]

[tex]E_{A}=\frac{9\times10^{9}\times 4.25 \times 10^{-6}}{2.42^2}[/tex]

EA = 6531.32 N/C

Electric field at B due to charge at C

[tex]E_{C}=\frac{Kq}{a^2}[/tex]

[tex]E_{C}=\frac{9\times10^{9}\times 4.25 \times 10^{-6}}{2.42^2}[/tex]

Ec = 6531.32 N/C

Electric field at B due to charge at D

[tex]E_{D}=\frac{Kq}{2a^2}[/tex]

[tex]E_{D}=\frac{9\times10^{9}\times 4.25 \times 10^{-6}}{2\times 2.42^2}[/tex]

ED = 3265.66 N/C

Now resolve the components along X axis and Y axis

Ex = EA + ED Cos 45 = 6531.32 + 3265.66 x 0.707 = 8840.5 N/C

Ey = Ec + ED Sin 45 = 6531.32 + 3265.66 x 0.707 = 8840.5 N/C

The resultant electric field at B is given by

[tex]E=\sqrt{E_{x}^{2}+E_{y}^{2}}[/tex]

[tex]E=\sqrt{8840.5^{2}+8840.5^{2}}[/tex]

E = 12500 N/C

Explanation:

Answer:

[tex]1.250\times 10^4\ N/C.[/tex]

Explanation:

Given:

According to the Coulomb's law, the strength of the electric field at a point due to a charge [tex]q[/tex] at a point [tex]r[/tex] distance away is given by

[tex] E = \dfrac{kq}{r^2}[/tex]

where,

[tex]k[/tex] = Coulomb's constant = [tex]9\times 10^9\ Nm^2/C^2[/tex].

The direction of the electric field is along the line joining the point an d the charge.

The electric field at the point P due to charge at A is given by

[tex]E_A = \dfrac{kq}{a^2}[/tex]

Since, this electric field is along positive x axis direction, therefore,

[tex]\vec E_A = \dfrac{kq}{a^2}\ \hat i.[/tex]

[tex]\hat i[/tex] is the unit vector along the positive x-axis direction.

The electric field at the point P due to charge at B is given by

[tex]E_B = \dfrac{kq}{a^2}[/tex]

Since, this electric field is along negative y axis direction, therefore,

[tex]\vec E_B = \dfrac{kq}{a^2}\ (-\hat j).[/tex]

[tex]\hat j[/tex] is the unit vector along the positive y-axis direction.

The electric field at the point P due to charge at C is given by

[tex]E_C = \dfrac{kq}{r^2}[/tex]

where, [tex]r=\sqrt{a^2+a^2}=a\sqrt 2[/tex].

Since, this electric field is along the direction, which is making an angle of [tex]45^\circ[/tex] below the positive x-axis direction, therefore, the direction of this electric field is given by [tex]\cos(45^\circ)\hat i+\sin(45^\circ)(-\hat j)[/tex].

[tex]\vec E_C = \dfrac{kq}{2a^2}\ (\cos(45^\circ)\hat i+\sin(45^\circ)(-\hat j))\\=\dfrac{kq}{2a^2}\ (\dfrac{1}{\sqrt 2}\hat i-\dfrac{1}{\sqrt 2}\hat j)\\=\dfrac{kq}{2a^2}\ \dfrac{1}{\sqrt 2}(\hat i-\hat j).\\[/tex]

Thus, the total electric field at the point P is given by

[tex]\vec E = \vec E_A+\vec E_B +\vec E_C\\=\dfrac{kq}{a^2}\hat i+\dfrac{kq}{a^2}(-\hat j)+\dfrac{kq}{2a^2}\ \dfrac{1}{\sqrt 2}(\hat i-\hat j).\\=\left ( \dfrac{kq}{a^2}+\dfrac{kq}{2\sqrt 2\ a^2} \right )\hat i+\left (\dfrac{kq}{a^2}+\dfrac{kq}{2\sqrt 2\ a^2} \right )(-\hat j)\\=\dfrac{kq}{a^2}\left [\left ( 1+\dfrac{1}{2\sqrt 2} \right )\hat i+\left (1+\dfrac{1}{2\sqrt 2}  \right )(-\hat j)\right ]\\=\dfrac{kq}{a^2}(1.353\hat i-1.353\hat j)[/tex]

[tex]=\dfrac{(9\times 10^9)\times (4.25\times 10^{-6})}{2.42^2}\times (1.353\hat i-1.353\hat j)\\=(8.837\times 10^3\hat i-8.837\times 10^3\hat j)\ N/C.[/tex]

The magnitude of the electric field at the given point due to all the three charges is given by

[tex]E=\sqrt{(8.837\times 10^3)^2+(-8.837\times 10^3)^2}=1.250\times 10^4\ N/C.[/tex]

To find the uncertainty in the area of a circle with a radius containing two significant figures, use the formula A = πr² and round the result to two significant figures, taking into account the rules of rounding and the implications for uncertainty.

The question pertains to the calculation of the uncertainty in the area of a circle when provided with the radius which has a certain number of significant figures. Given the radius (r) of a circle as 4.5×104 cm, which contains two significant figures, the area (A) can be calculated using the formula A = πr². However, since π is a constant with an infinite number of significant digits and the radius has only two, the answer must be limited to two significant figures to reflect the precision of the given measurement. Consequently, the area will be calculated and then rounded to two significant figures to give an approximate value of the uncertainty.

Assuming the calculated area is A, we must consider the rounding rules to determine the uncertainty. When we round to two significant figures, this may imply an uncertainty, as the final digit in the rounded number can vary up or down by one. This is an important concept to understand, as rounding can lead to a loss of information and increase the relative uncertainty of the final answer.

Answer:

1.81 x 10^-4 m/s

Explanation:

M = 98700 kg

m = 780 kg

d = 201 m

Let the speed of second asteroid is v.

The gravitational force between the two asteroids is balanced by the centripetal force on the second asteroid.

[tex]\frac{GMm}{d^{2}}=\frac{mv^2}{d}[/tex]

[tex]v=\sqrt{\frac{GM}{d}}[/tex]

Where, G be the universal gravitational constant.

G = 6.67 x 10^-11 Nm^2/kg^2

[tex]v=\sqrt{\frac{6.67 \times 10^{-11}\times 98700}{201}}[/tex]

v = 1.81 x 10^-4 m/s

Answer:

frequency = 19.56 Hz

Explanation:

given data

length L = 7 m

mass m = 7 g

mass M = 2.50 kg

distance d = 4 m

to find out

fundamental frequency

solution

we know here frequency formula is

frequency = [tex]\frac{v}{2d}[/tex]   ...........1

so here d is given = 4

and v = [tex]\sqrt{\frac{T}{\mu} }[/tex]  ..........2

tension T = Mg = 2.50 × 9.8 = 24.5 N

and μ = [tex]\frac{m}{l}[/tex] =  [tex]\frac{7*10^{-3} }{7}[/tex] = [tex]10^{-3}[/tex] kg/m

so from equation 2

v = [tex]\sqrt{\frac{24.5}{10^{-3}} }[/tex]

v = 156.52

and from equation 1

frequency = [tex]\frac{v}{2d}[/tex]

frequency = [tex]\frac{156.52}{2(4)}[/tex]

frequency = 19.56 Hz

Final answer:

The fundamental frequency of the vibrating cord is approximately 14.18 Hz.

Explanation:

To determine the fundamental frequency of the vibrating cord, we can use the equation for the fundamental frequency of a vibrating string:

f1 = 1/2L * sqrt(T / μ)

Where f1 is the fundamental frequency, L is the total length of the cord, T is the tension in the cord, and μ is the linear density of the cord.

Plugging in the given values, we have: f1 = 1/2 * 7.00 * sqrt(90 / 0.007)

Simplifying this equation gives us the fundamental frequency of the cord: f1 ≈ 14.18 Hz

Answer:

Explanation:

mass of astronaut, m = 60 kg

mass of space craft, M = 120 kg

t = 1.5 s

Force on space craft = 30 N

(a) According to Newton's third law

Force on spacecraft by the astronaut = Force on astronaut by the space craft

Force on astronaut by the space craft  = 30 N

(b) According to Newton's second law

Force  = mass x acceleration

Let a be the acceleration of the astronaut

30 = 60 x a

a = 0.5 m/s^2

Let A be the acceleration of the spacecraft

30 = 120 x A

a = 0.25 m/s^2

Answer:4.95 m/s

Explanation:

Given

Ball rises to a height of 1.25 m

Let u be the velocity of player while leaving ground .

Its final velocity will be zero as he reaches a height of 1.25 m

thus

[tex]u^2=2gh[/tex]

[tex]u=\sqrt{2\times 9.81\times 1.25}[/tex]

[tex]u=\sqrt{24.525}[/tex]

u=4.952 m/s

Final answer:

To reach a height of 1.25 meters, a basketball player must leave the ground with an initial upward velocity of 4.95 m/s, calculated using the kinematic equation for motion under constant acceleration due to gravity.

Explanation:

To find the required initial velocity, we use the kinematic equation which relates distance (d), initial velocity ([tex]v_i[/tex] ), final velocity ([tex]v_f[/tex]), acceleration (a), and time (t). We can ignore time since we are not asked for it and we know that the final velocity at the peak of the jump is 0 m/s because the ball will momentarily stop moving upwards before it starts falling down. We will use the acceleration due to gravity, which is approximately 9.81 m/s².

The equation is:

[tex]v_f^2 = v_i^2 + 2ad[/tex]

Setting [tex]v_f[/tex] to 0 m/s (since the ball stops at the peak), the equation simplifies to:

[tex]v_i[/tex] = √(2ad)

Plugging in the values we get:

[tex]v_i[/tex]  = √(2 * 9.81 m/s² * 1.25 m) = √(24.525 m²/s²)

[tex]v_i[/tex]  = 4.95 m/s

Therefore, the basketball player must leave the ground with an initial velocity of 4.95 m/s to reach a height of 1.25 meters.

Answer:

Answer:

energy

Explanation:

The capacity to do work is called energy.

There are several types of energy. Some of the types of energy are given below:

1. Kinetic energy

2. Potential energy

3. Solar energy

4. thermal energy

5. Sound energy

6. Nuclear energy, etc.

The SI unit of energy is Joule.

Some other units of energy are

erg, eV, etc.

1 J = 10^7 erg

1 eV = 1.6 x 10^-19 J

The commercial unit of energy is kilowatt hour.

1 Kilo watt hour = 3.6 x 10^6 J

Explanation:

Answer:D-velocity

Explanation:

Given

Constant linear acceleration i.e. acceleration is constant

a=constant

and we know [tex]\frac{\mathrm{d}V}{\mathrm{d} t}=a[/tex]

Thus velocity is changing uniformly to give constant acceleration

While displacement and distance are function of time thus they are not changing uniformly.

During constant linear acceleration, the following quantity changes uniformly: velocity. Therefore option D is correct.

Constant linear acceleration refers to a situation where an object's acceleration remains constant over a specific period of time. In this case, the rate of change of velocity is uniform, meaning the velocity increases or decreases by the same amount in equal time intervals.

Velocity is the rate of change of displacement and is directly affected by acceleration. During constant linear acceleration, the velocity changes uniformly. It either increases or decreases at a constant rate, depending on the direction of acceleration.

Therefore,

During constant linear acceleration, the following quantity changes uniformly: velocity.

Know more about linear acceleration:

https://brainly.com/question/13385172

#SPJ6

Answer:

Density of the swimmer = [tex]0.0342\ lbm/in^3[/tex].

Explanation:

Assuming,

Given:

The density of an object is defined as the mass of the object per unit volume.

Therefore,

[tex]\rho =\dfrac{m}{V}\ \Rightarrow m = \rho V\ \ .........\ (1).[/tex]

Since only 95% of the body of the swimmer is inside the water, therefore,

[tex]V_w = 95\%\ \text{of}\ V=\dfrac{95}{100}\times V = 0.95V.[/tex]

According to Archimedes' principle,

[tex]m=m_w\\[/tex]

Using (1),

[tex]\rho V=\rho_w V_w\\\rho V = 0.036\ lbm/in^3\times 0.95 V\\\rho=0.036\times 0.95\ lbm/in^3=0.0342\ lbm/in^3.[/tex]

Answer:

electron moving at 91.82 m/s

Explanation:

given data

distance d1 = 6 cm

distance d2 = 3 cm

to find out

how fast will the electron be moving

solution

we know potential energy formula that is

potential energy = [tex]k\frac{q1q2}{r}[/tex]

here k is 9× [tex]10^{9}[/tex] N-m²/C²

m mass of electron is 9.11 × [tex]10^{-31}[/tex] kg

and q = 1.60 × [tex]10^{-19}[/tex] C

we consider here initial potential energy = U1

U1 = [tex]9*10^{9}\frac{1.60*10^{-19}*1.60*10^{-19}}{0.06^2}[/tex]

U1 = 3.84 × [tex]10^{-27}[/tex] J

and final potential energy = U2

U2 = [tex]9*10^{9}\frac{1.60*10^{-19}*1.60*10^{-19}}{0.03}[/tex]

U2 = 7.68 × [tex]10^{-27}[/tex] J

and speed of electron = v

so we will apply here conservation of energy

0.5×m×v² = U2 - U1    ................1

so

0.5×9.11× [tex]10^{-31}[/tex] ×v² =  3.84 × [tex]10^{-27}[/tex]

v = 91.82 m/s

so electron moving at 91.82 m/s

Answer:

The speed of the electron is 91.86 m/s.

Explanation:

Given that,

Distance of electron from proton  = 6.00 cm

Distance of proton = 3.00 cm

We need to calculate the initial potential energy

[tex]U_{1}=\dfrac{kq_{1}q_{2}}{r}[/tex]

Put the value into the formula

[tex]U_{i}=\dfrac{9\times10^{9}\times(1.6\times10^{-19})^2}{6.00\times10^{-2}}[/tex]

[tex]U_{i}=3.84\times10^{-27}\ J[/tex]

The final potential energy

[tex]U_{f}=\dfrac{9\times10^{9}\times(1.6\times10^{-19})^2}{3.00\times10^{-2}}[/tex]

[tex]U_{f}=7.68\times10^{-27}\ J[/tex]

We need to calculate the speed of the electron

Using conservation of energy

[tex]\dfrac{1}{2}mv^2=U_{f}-U_{i}[/tex]

Put the value into the formula

[tex]\dfrac{1}{2}\times9.1\times10^{-31}\times v^2=7.68\times10^{-27}-3.84\times10^{-27}[/tex]

[tex]v=\sqrt{\dfrac{2\times3.84\times10^{-27}}{9.1\times10^{-31}}}[/tex]

[tex]v=91.86\ m/s[/tex]

Hence, The speed of the electron is 91.86 m/s.

Answer:

The magnitude of charge on each is [tex]5.707\times 10^{13} C[/tex]

Solution:

As per the question:

Mass of Earth, [tex]M_{E} = 5.98\times 10^{24} kg[/tex]

Mass of Moon, [tex]M_{M} = 7.35\times 10^{22} kg[/tex]

Now,

The gravitational force of attraction between the earth and the moon, if 'd' be the separation distance between them is:

[tex]F_{G} = \frac{GM_{E}M_{M}}{d^{2}}[/tex]        (1)

Now,

If an identical charge 'Q' be placed on each, then the Electro static repulsive force is given by:

[tex]F_{E} = \frac{1}{4\pi\epsilon_{o}}\frac{Q^{2}}{d^{2}}[/tex]           (2)

Now, when the net gravitational force is zero, the both the gravitational force and electro static force mut be equal:

Equating eqn (1) and (2):

[tex]\frac{GM_{E}M_{M}}{d^{2}} = \frac{1}{4\pi\epsilon_{o}}\frac{Q^{2}}{d^{2}}[/tex]

[tex](6.67\times 10^{- 11})\times (5.98\times 10^{24})\times (7.35\times 10^{22}) = (9\times 10^{9}){Q^{2}}[/tex]

[tex]\sqrt{\farc{(6.67\times 10^{- 11})\times (5.98\times 10^{24})\times (7.35\times 10^{22})}{9\times 10^{9}}} = Q[/tex]

Q = [tex]\pm 5.707\times 10^{13} C[/tex]

Hi!

Letw call A to the initially neutral rod, and B to the positively charged. When they are close to each other, the positive charges in B attract the negative charges in A, and repelle the positive ones. If you ground A, negative charges from ground (your body, in this case), flow to A attracted by the positive charges in B, and positive charges in A flow to ground, so finally A results negatively charged

Answer:

Negative charges

Explanation:

The procedure described above is known in physics as charging by electrostatic induction. If we desire to impart negative charges to a hitherto neutral rod, we bring a positively charged rod near it without allowing the two insulated rods to touch each other. If the neutral rod is earthed, negative charges remain on the rod.

Answer:

17.64 km/h

Explanation:

mass of car, m = 1000 kg

Kinetic energy of car, K = 1.2 x 10^4 J

Let the speed of car is v.

Use the formula for kinetic energy.

[tex]K = \frac{1}{2}mv^{2}[/tex]

By substituting the values

[tex]1.2\times 10^{4} = \frac{1}{2}\times 1000\times v^{2}[/tex]

v = 4.9 m/s

Now convert metre per second into km / h

We know that

1 km = 1000 m

1 h = 3600 second

So, [tex]v = \left (\frac{4.9}{1000}   \right )\times \left ( \frac{3600}{1} \right )[/tex]

v = 17.64 km/h

Thus, the reading of speedometer is 17.64 km/h.

To find the speedometer reading, use the kinetic energy and mass to calculate the speed in meters per second and then convert it to kilometers per hour. The calculated speed is approximately 88 km/hr.

To determine the speedometer reading of the car in km/hr based on the kinetic energy given, we use the kinetic energy formula KE = ½ mv². We rearrange this formula to solve for speed (v): v = √(2KE/m).

Plugging in the provided kinetic energy of 1.2 x 10´ J and the car's mass of 1000 kg, we get:

v = √(2 × 1.2 x 10´ J / 1000 kg)

v = √(24 x 10´ J/kg)

v ≈ 24.49 m/s

To convert from m/s to km/hr, we multiply by 3.6:

v = 24.49 m/s × 3.6 km/hr per m/s ≈ 88.16 km/hr

Therefore, the speedometer would read approximately 88 km/hr.

Answer:4.93 m/s

Explanation:

Given

height to reach is (h )1.24 m

here Let initial velocity is u

using equation of motion

[tex]v^2-u^2=2ah[/tex]

here Final Velocity v=0

a=acceleration due to gravity

[tex]0-u^2=2\left ( -g\right )h[/tex]

[tex]u=\sqrt{2gh}[/tex]

[tex]u=\sqrt{2\times 9.81\times 1.24}[/tex]

[tex]u=\sqrt{24.328}[/tex]

u=4.93 m/s

Answer:

Because by potentiometer we can drop voltage according to our requirement but in set of two series resistors we can not do this.

Explanation:

As we know that the potentiometer can act as a variable resistor but resistors have constant value.

In any circuit if we insert resistors they will drop a constant value if we insert a potentiometer we will drop the value of voltage by according to our will.

Therefore, by the above discussion it can be say that the potentiometer is better than two resesters in series.

Answer:

For 0.24 sec the person was in the air.

Explanation:

Given that,

Height = 1 m

Initial velocity = 3 m/s

We need to calculate the time

Using equation of motion

[tex]s=ut+\dfrac{1}{2}gt^2[/tex]

Where, u = initial velocity

s = height

Put the value into the formula

[tex]1 =3\times t+\dfrac{1}{2}\times9.8\times t^2[/tex]

[tex]4.9t^2+3t-1=0[/tex]

[tex]t = 0.24\ sec[/tex]

On neglecting the negative value of time

Hence, For 0.24 sec the person was in the air.