3 years ago

Find the speed required to throw a ball straight up and it return 6 seconds later. Neglect air resistance

Physics

1 answer:

Nadya [2.5K]3 years ago

8 0

Answer:

the ball will go up 3s and down 3s

v=gt

where t=3s and g=9.8m/s^2

distance=v0(t)+(1/2)gt^2

where initial velocity (v0)=0

Explanation:

You might be interested in

As the ambulance got closer, Marge noticed that the pitch of the siren got higher. This happened because

AfilCa [17]

A) the sound waves were pushed closer together

6 0

3 years ago

Read 2 more answers

Calculate the phase angle (in radians) for a circuit with a maximum voltage of 12 V and w-50 Hz. The voltage source is connected

Vinvika [58]

Answer:

The phase angle is 0.0180 rad.

(c) is correct option.

Explanation:

Given that,

Voltage = 12 V

Angular velocity = 50 Hz

Capacitance $C= 20\times10^{-2}\ F$

Inductance $L=20\times10^{-3}\ H$

Resistance $R= 50\ Omega$

We need to calculate the impedance

Using formula of impedance

$z=\sqrt{R^2+(\omega L-\dfrac{1}{\omega C})^2}$

$z=\sqrt{50^2+(50\times20\times10^{-3}-\dfrac{1}{50\times20\times10^{-2}})^2}$

$z=50.00$

We need to calculate the phase angle

Using formula of phase angle

$\theta=\cos^{-1}(\dfrac{R}{z})$

$\theta=\cos^{-1}(\dfrac{50}{50.00})$

$\theta=0.0180\ rad$

Hence, The phase angle is 0.0180 rad.

3 0

3 years ago

A change in position of an object relating to time

MaRussiya [10]

The answer is: Motion!

Have a great day

3 0

3 years ago

Read 2 more answers

A 53.3 kg woman slides down a 35.0° hill with an acceleration of 4.10 m/s. What is the friction force acting on the woman?

lorasvet [3.4K]

Answer:

I attached an image that should help.

Explanation:

Check it out.

5 0

3 years ago

A test charge of -1.4 x 10-7 coulombs experiences a force of 5.4 x 10-1 newtons. Calculate the magnitude of the electric field c

VARVARA [1.3K]

Answer:

3.86×10⁶ Newton/coulombs

Explaination:

Applying,

E = F/q....................... Equation 1

Where E = Electric Field, F = Force, q = charge.

From the question,

Given: F = 5.4×10⁻¹ N, q = -1.4×10⁻⁷ coulombs

Substitute these values into equation 1

E = 5.4×10⁻¹/ -1.4×10⁻⁷

E = -3.86×10⁶ Newtons/coulombs

Hence the magnitude of the electric field created by the

negative test charge is 3.86×10⁶ Newton/coulombs

5 0

3 years ago