If the sun, Earth and moon are lined up as shown as above, the Earth

You throw a rock straight up and find that it returns to your hand 3.40 s after it left your hand. Neglect air resistance. What

Soloha48 [4]

Answer:

The maximum height of the rock is 14.2 m

Explanation:

The equations that describe the height and velocity of the rock are the following:

y = y0 + v0 · t + 1/2 · g · t²

v = v0 + g · t

Where:

y = height of the object at time t

y0 = initial height

t = time

g = acceleration due to gravity (-9.8 m/s² if upward is positive)

v = velocity of the object at time t

We know that at t = 3.40 s, the rock is in your hand again. Then, if we place the origin of the frame of reference at your hand, the position of the rock at 3.40 s is 0 m. Using the equation of the position, we can calculate the initial velocity that we will need to obtain the max-height.

y = y0 + v0 · t + 1/2 · g · t²

0 = v0 · 3.40 s - 1/2 · 9.8 m/s² · (3.40 s)²

(1/2 · 9.8 m/s² · (3.40 s)² ) / 3.40 s = v0

v0 = 16.7 m/s

At max-height, the velocity of the rock is 0. Then, using the equation of velocity we can calculate the time it takes the rock to reach the max-height. With that time, we can calculate the maximum height.

v = v0 + g · t (at max-height, v = 0)

0 = 16.7 m/s - 9.8 m/s² · t

- 16.7 m/s / - 9.8 m/s² = t

t = 1.70 s

Now, using this time in the equation of height:

y = y0 + v0 · t + 1/2 · g · t²

y = 0 m + 16.7 m/s · 1.70 s - 1/2 · 9.8 m/s² · (1.70 s)²

y = 14.2 m

The maximum height of the rock is 14.2 m

8 0

3 years ago

A coil has N turns enclosing an area of A. In a physics laboratory experiment, the coil is rotated during the time interval Δt f

scoray [572]

Answer:

BA

0

$\dfrac{NBA}{\Delta t}$

Explanation:

B = Magnetic field

A = Area

$\theta$ = Angle

t = Time taken

Before rotation the magnetic flux is given by

$\phi_i=BAsin\theta\\\Rightarrow \phi_i=BAsin90\\\Rightarrow \phi_i=BA$

Magnetic flux is BA

After rotation the magnetic flux is given by

$\phi_i=BAsin\theta\\\Rightarrow \phi_i=BAsin0\\\Rightarrow \phi_i=0$

The magnetic flux is 0

Magnitude of emf is given by

$\epsilon=NA\dfrac{d\phi}{dt}\\\Rightarrow \epsilon=\dfrac{NBA}{\Delta t}$

The magnitude of the average emf induced in the entire coil is $\dfrac{NBA}{\Delta t}$

4 0

3 years ago

The slope of the line tangent to the curve on a position-time graph at a specific time is the

Rudik [331]

Answer:

I do I make a brinliest can you please can me

7 0

3 years ago

A gymnast of mass 62.0 kg hangs from a vertical rope attached to the ceiling. You can ignore the weight of the rope and assume t

MrRissso [65]

Answer:

a) T = 608.22 N

b) T = 608.22 N

c) T = 682.62 N

d) T = 533.82 N

Explanation:

Given that the mass of gymnast is m = 62.0 kg

Acceleration due to gravity is g = 9.81 m/s²

Thus; The weight of the gymnast is acting downwards and tension in the string acting upwards.

So;

To calculate the tension T in the rope if the gymnast hangs motionless on the rope; we have;

T = mg

= (62.0 kg)(9.81 m/s²)

= 608.22 N

When the gymnast climbs the rope at a constant rate tension in the string is

= (62.0 kg)(9.81 m/s²)

= 608.22 N

When the gymnast climbs up the rope with an upward acceleration of magnitude

a = 1.2 m/s²

the tension in the string is T - mg = ma (Since acceleration a is upwards)

T = ma + mg

= m (a + g )

= (62.0 kg)(9.81 m/s² + 1.2 m/s²)

= (62.0 kg) (11.01 m/s²)

= 682.62 N

When the gymnast climbs up the rope with an downward acceleration of magnitude

a = 1.2 m/s² the tension in the string is mg - T = ma (Since acceleration a is downwards)

T = mg - ma

= m (g - a )

= (62.0 kg)(9.81 m/s² - 1.2 m/s²)

= (62.0 kg)(8.61 m/s²)

= 533.82 N

5 0

2 years ago

To test the quality of a tennis ball, you drop it onto the floor from a height of 4.00 m. It rebounds to a height of 2.00 m. If

arlik [135]

Answer:

Part a)

$a = 1260.3 m/s^2$

Part b)

Direction = upwards

Explanation:

When ball is dropped from height h = 4.0 m

then the speed of the ball just before it will strike the ground is given as

$v_f^2 - v_i^2 = 2 a d$

$v_1^2 - 0^2 = 2(9.81)(4.0)$

$v_1 = 8.86 m/s$

Now ball will rebound to height h = 2.00 m

so the velocity of ball just after it will rebound is given as

$v_f^2 - v_i^2 = 2 a d$

$0 - v_2^2 = 2(-9.81)(2.00)$

$v_2 = 6.26 m/s$

Part a)

Average acceleration is given as

$a = \frac{v_f - v_i}{\Delta t}$

$a = \frac{6.26 - (-8.86)}{12.0 \times 10^{-3}}$

$a = 1260.35 m/s^2$

Part B)

As we know that ball rebounds upwards after collision while before collision it is moving downwards

So the direction of the acceleration is vertically upwards

7 0

3 years ago