PHYSICS COLLEGE

A tennis player hits a ball 2.0 m above the ground. The ball leaves his racquet with a speed of 20 m/s at an angle 5.0 ∘ above the horizontal. The horizontal distance to the net is 7.0 m, and the net is 1.0 m high. Does the ball clear the net?

Answers

Answer 1

Answer:

Answer:

ball clears the net

Explanation:

$v_(o)$ = initial speed of launch of the ball = 20 ms^{-1}

$\theta$ = angle of launch = 5 deg

Consider the motion of the ball along the horizontal direction

$v_(ox)$ = initial velocity = $v_(o) Cos\theta = 20 Cos5 = 19.92 ms^(-1)$

$t$ = time of travel

$X$ = horizontal displacement of the ball to reach the net = 7 m

Since there is no acceleration along the horizontal direction, we have

$X = v_(ox) t \n7 = v_(ox) t\nt = (7)/(v_(ox))$ Eq-1

Consider the motion of the ball along the vertical direction

$v_(oy)$ = initial velocity = $v_(o) Sin\theta = 20 Sin5 = 1.74 ms^(-1)$

$t$ = time of travel

$Y_(o)$ = Initial position of the ball at the time of launch = 2 m

$Y$ = Final position of the ball at time "t"

$a_(y)$ = acceleration in down direction = - 9.8 ms⁻²

Along the vertical direction , position at any time is given as

$Y = Y_(o) + v_(oy) t + (0.5) a_(y) t^(2)\nY = 2 + (20 Sin5) ((7)/(20 Cos5)) + (0.5) (- 9.8) ((7)/(20 Cos5))^(2)\nY = 2.00758 m\n$

Since Y > 1 m

hence the ball clears the net

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Question 8 of 10It takes a person 22 seconds to swim in a straight line from the south end of
a pool to the north end of the pool, a distance of 28 meters. What is the
swimmer's velocity?
A. 1.3 m/s south
B. 1.3 m/s north
C. 0.8 m/s south
D. 0.8 m/s north

Answers

The correct answer is B. 1.3 m/s north

Explanation:

Velocity is a factor that describes how fast or slow the motion of a body occurs and its direction. Moreover, this can be calculated by dividing the total displacement into the time of movement, and the final result is expressed in units such as meters per second followed by the direction, for example, 152 m/s south. The process to calculate the velocity of the swimmer is shown below.

$v = (d)/(t)$

$v = (28 meters)/(22 seconds)$

$v = 1.27 m/s$

This means the velocity of the swimmer is 1.27 m/s, which can be rounded as 1.3 m/s. Additionally, if the direction is considered it would be 1.3 m/s north because the swimmer went from the south of the pool to its north.

Answer:

the answer is B

Explanation:

confirmed

3. A particle of charge +7.5 µC is released from rest at the point x = 60 cm on an x-axis. The particle begins to move due to the presence of a charge ???? that remains fixed at the origin. What is the kinetic energy of the particle at the instant it has moved 40 cm if a) ???? = +20 µC and b) ???? = −20 µC?

Answers

Answer:

HSBC keen vs kg get it yyyyyuuy

Explanation:

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A 30 kg child on a 2 m long swing is released from rest when the swing supports make an angle of 34 ◦ with the vertical. The acceleration of gravity is 9.8 m/s 2 . If the speed of the child at the lowest position is 2.31547 m/s, what is the mechanical energy dissipated by the various resistive

Answers

Answer:

Energy dissipated = 13.453 Joules

Explanation:

In order to solve this problem, we first compute the gravitational potential energy the child has, and then find the kinetic energy at the lowest position.

The gravitational potential energy (relative to lowest position) is found as follows:

G.P.E = mass * gravity * height

Where, Height = 2 - 2 * Cos(34°)

Height = 0.3193 meters

G.P.E = 30 * 9.8 * 0.3193

G.P.E = 93.874 J

Kinetic energy:

K.E = 0.5 * mass * velocity^2

K.E = 0.5 * 30 * 2.31547^2

K.E = 80.421 J

Energy dissipated = G.P.E - K.E

Energy dissipated = 93.874 - 80.421

Energy dissipated = 13.453 J

Why Coulomb force is called "Mutual Force"???????????????????????????????????????

Answers

Answer

The Columb's law is the same as Gravitational law

Explanation

As we see the formula of both Coulomb and Gravitational Law,

$F_g = GM_1M_2/R_2$ (1)

$F_c = kq_1q_2/r_2$ (2)

The masses (M) in formula (1) experiencing the force of gravitational pull with each other which varies with changing the distance. In the formula (2), the charges also are felling the forces on each other which varies with distance. The charges and masses are just like the objects which are experiencing the forces which have a common factor as distance. The gravitational force is also called the mutual forces.

A 500 W heating coil designed to operate from 110 V is made of Nichrome 0.500 mm in diametera.Assuming the resistivity of the nichrome remains constant at is 20.0 degrees C value find the length of wire used.b. Now consider the variation of resistivity with temperature. What power is delivered to the coil of part (a) when it is warmed to 1200 degrees C.?

Answers

(a) Length of the wire is 3.162 m

(b)Power delivered to the coil is 339.7 W

Electrical Power:

The electrical power is given by

P = V² / R

R = V² / P

Resistance of the heating coil, R

R = (110² / 500)

R = 12100 / 500

R = 24.2 Ω

Now the resistivity of a wire is given by

ρ= RA/L

here ρ = 1.50×10⁻⁶ Ωm

so after rearranging we get:

L = RA / ρ

Now, the radius of wirer = 0.5 / 2 mm = 0.25 mm = 2.5×10⁻⁴ m

So the cross sectional area can be calculated as follows

$A = \pi r^2\n\nA = \pi * (2.5*10^(-4))^2\n\nA = 1.96*10^(-7) m^2$

hence,

$L = (24.2 *1.96*10^(-7) / 1.50*10^(-6)) \n\nL = 3.162\; m$

(b)The dependency of resistance with temperature is as follows:

R = R₀[1 + αΔT]

α = $4*10^(-4)^\;oC^(-1)$ for Nichrome

$R' = R [1 + \alpha (1200 - 20) ]\n\nR' = R[1 + \alpha (1180) ]\n\nR' = 24.2[ 1 + 4*10^(-4) * 1180 ]\n\nR' = 24.2[1 + 0.472]\n\nR' = 24.2 * 1.472\n\nR' = 35.62 \;\Omega$

So the power generated is :

P = V² / R

P = (110² / 35.62)

P = 12100/ 35.62

P = 339.70 watts

Learn more about electrical power:

brainly.com/question/26174188

Answer:

a) 3.162 m

b) 339.7 W

Explanation:

Assume ρ = 1.50*10^-6 Ωm, and

α = 4.000 10-4(°C)−1 for Nichrome

To solve this, we would use the formula

P = V² / R

So when we rearrange and make R subject of formula, we have

R = V² / P

Resistance of the heating coil, R

R = (110² / 500)

R = 12100 / 500

R = 24.2 ohms

Recall the formula for resistivity of a wire

R = ρ.L/A

Again, in rearranging and making L subject of formula, we have

L = R.A / ρ

To make it uniform, we convert our radius from mm to m.

Diameter, D = 0.5 mm

Radius of wire = 0.5 / 2 mm = 0.25 mm = 0.00025 m

We then use this radius to find our area

A = πr²

A = π * 0.00025²

A = 1.96*10^-7 m²

And finally, we solve for L

L = (24.2 * 1.96*10^-7 / 1.50*10^-6) =

L = 3.162 m

(b)

Temperature coefficient of resistance.

R₁₂₀₀ = R₂₀[1 + α(1200 - 20.0) ]

R₁₂₀₀ = R₂₀[1 + α(1180) ]

R₁₂₀₀ = 24.2[ 1 + 4.*10^-4 * 1180 ]

R₁₂₀₀ = 24.2[1 + 0.472]

R₁₂₀₀ = 24.2 * 1.472

R₁₂₀₀ = 35.62 ohms

Putting this value of R in the first formula from part a, we have

P = V² / R

P = (110² / 35.62)

P = 12100/ 35.62

P = 339.70 watts

Certain insects can achieve seemingly impossible accelerations while jumping. the click beetle accelerates at an astonishing 400g over a distance of 0.52 cm as it rapidly bends its thorax, making the "click" that gives it its name. part a assuming the beetle jumps straight up, at what speed does it leave the ground? part b how much time is required for the beetle to reach this speed? part c ignoring air resistance, how high would it go?

Answers

(a) The launching velocity of the beetle is 6.4 m/s

(b) The time taken to achieve the speed for launch is 1.63 ms

(a) The beetle starts from rest and accelerates with an upward acceleration of 400 g and reaches its launching speed in a distance 0.53 cm. Here g is the acceleration due to gravity.

Use the equation of motion,

$v^2=u^2+2as$

Here, the initial velocity of the beetle is u, its final velocity is v, the acceleration of the beetle is a, and the beetle accelerates over a distance s.

Substitute 0 m/s for u, 400 g for a, 9.8 m/s² for g and 0.52×10⁻²m for s.

$v^2=u^2+2as\n = (0 m/s)^2+2 (400)(9.8 m/s^2)(0.52*10^-^2 m)\n =40.768 (m/s)^2\n v=6.385 m/s$

The launching speed of the beetle is 6.4 m/s.

(b) To determine the time t taken by the beetle for launching itself upwards is determined by using the equation of motion,

$v=u+at$

Substitute 0 m/s for u, 400 g for a, 9.8 m/s² for g and 6.385 m/s for v.

$v=u+at\n 6.385 m/s = (0 m/s) +400(9.8 m/s^2)t\n t = (6.385 m/s)/(3920 m/s^2) = 1.63*10^-^3s=1.63 ms$

The time taken by the beetle to launch itself upwards is 1.62 ms.

(c) After the beetle launches itself upwards, it is acted upon by the earth's gravitational force, which pulls it downwards towards the earth with an acceleration equal to the acceleration due to gravity g. Its velocity reduces and when it reaches the maximum height in its path upwards, its final velocity becomes equal to zero.

Use the equation of motion,

$v^2=u^2+2as$

Substitute 6.385 m/s for u, -9.8 m/s² for g and 0 m/s for v.

$v^2=u^2+2as\n (0m/s)^2=(6.385 m/s)^2+2(-9.8m/s^2)s\n s=((6.385 m/s)^2)/(2(9.8m/s^2)) =2.08 m$

The beetle can jump to a height of 2.1 m

We have that for the Question the Speed,Time and Height are

u=6.38m/s
T=13sec
h=2m

From the question we are told

Certain insects can achieve seemingly impossible accelerations while jumping.

the click beetle accelerates at an astonishing 400g over a distance of 0.52 cm as it rapidly bends its thorax, making the "click" that gives it its name.