Ask question

Ask question

All categories

4 years ago

7

A basketball is shot from 2 meters up at an angle of 60° above the x axis at an initial velocity of 9 m/s. What is the maximum h

eight the ball will reach?

2 answers:

Nikitich [7]4 years ago

7 0

Hope this helps a little

initial distance up = 2

initial velocity component up = 9 sin 60 = 7.79

v = 9 sin 60 - 9.8 t

when v = 0, we are there

9.8 t = 7.79

t = .795 seconds to top

h = 2 + 7.79(.795) - 4.9(.795^2)

nika2105 [10]4 years ago

5 0

Answer:

6.2m

Explanation:

Maximum height is the height reached by an object launched into space before falling under the influence of gravity. Mathematically,

Maximum height (H) = u²sin²(theta)/2g

Given initial velocity (u) = 9m/s

Angle of launch = 60°

Acceleration due to gravity g = 9.8m/s

H = 9²(sin60)²/9.8

H = 81(0.8660)²/9.8

H = 6.2m

The maximum height reached by the ball is 6.2m

You might be interested in

A charge of Q is fixed in space. A second charge of q was first placed at a distance r1 away from Q. Then it was moved along a s

topjm [15]

Answer:

$\Delta U = \frac{Qq}{4\pi\epsilon_0}(\frac{1}{r_2^2}-\frac{1}{r_1^2})$

Explanation:

The electrostatic potential energy is given by the following formula

$U = \frac{1}{4\pi\epsilon_0}\frac{q_1q_2}{r^2}$

Now, we will apply this formula to both cases:

$U_1 = \frac{1}{4\pi\epsilon_0}\frac{Qq}{r_1^2}\\U_2 = \frac{1}{4\pi\epsilon_0}\frac{Qq}{r_2^2}$

So, the change in the potential energy is

$\Delta U = U_2 - U_1 = \frac{Qq}{4\pi\epsilon_0}(\frac{1}{r_2^2}-\frac{1}{r_1^2})$

7 0

3 years ago

2. What is the power rating of an engine capable of lifting a 100 kg object 5 m vertically

Leya [2.2K]

Work=f.d
Work=100*50 = 500
Power = work/time = 500/4
=125 watt

7 0

3 years ago

A spring stretches 0.2 cm per newton of applied force. An object is suspended from the spring and a deflection of 3 cm is observ

siniylev [52]

Answer:

$m=1.53kg$

Explanation:

To solve this problem we use the Hooke's Law:

$F=k*\Delta x$ (1)

F is the Force needed to expand or compress the spring by a distance Δx.

The spring stretches 0.2cm per Newton, in other words:

1N=k*0.2cm ⇒ k=1N/0.2cm=5N/cm

The force applied is due to the weight

$F=mg$

We replace in (1):

$mg=k*\Delta x$

We solve the equation for m:

$m=k*\Delta x/g=5*3/9.81=1.53kg$

7 0

3 years ago

A merry-go-round with a rotational inertia of 600 kg m2 and a radius of 3.0 m is initially at rest. A 20 kg boy approaches the m

nekit [7.7K]

Answer:

The velocity of the merry-go-round after the boy hops on the merry-go-round is 1.5 m/s

Explanation:

The rotational inertia of the merry-go-round = 600 kg·m²

The radius of the merry-go-round = 3.0 m

The mass of the boy = 20 kg

The speed with which the boy approaches the merry-go-round = 5.0 m/s

$F_T \cdot r = I \cdot \alpha = m \cdot r^2 \cdot \alpha$

Where;

$F_T$ = The tangential force

I = The rotational inertia

m = The mass

α = The angular acceleration

r = The radius of the merry-go-round

For the merry go round, we have;

$I_m \cdot \alpha_m = I_m \cdot \dfrac{v_m}{r \cdot t}$

$I_m$ = The rotational inertia of the merry-go-round

$\alpha _m$ = The angular acceleration of the merry-go-round

$v _m$ = The linear velocity of the merry-go-round

t = The time of motion

For the boy, we have;

$I_b \cdot \alpha_b = m_b \cdot r^2 \cdot \dfrac{v_b}{r \cdot t}$

Where;

$I_b$ = The rotational inertia of the boy

$\alpha _b$ = The angular acceleration of the boy

$v _b$ = The linear velocity of the boy

t = The time of motion

When the boy jumps on the merry-go-round, we have;

$I_m \cdot \dfrac{v_m}{r \cdot t} = m_b \cdot r^2 \cdot \dfrac{v_b}{r \cdot t}$

Which gives;

$v_m = \dfrac{m_b \cdot r^2 \cdot \dfrac{v_b}{r \cdot t} \cdot r \cdot t}{I_m} = \dfrac{m_b \cdot r^2 \cdot v_b}{I_m}$

From which we have;

$v_m = \dfrac{20 \times 3^2 \times 5}{600} = 1.5$

The velocity of the merry-go-round, $v_m$ , after the boy hops on the merry-go-round = 1.5 m/s.

5 0

3 years ago

If toner particles in a laser printer have a negative charge, then what charge do you think the surface of the paper in the prin

Elina [12.6K]

Answer:

Explanation:

As it moves along, the paper is given a strong negative electrical charge by another corona wire. When the paper moves near the drum, its negative charge attracts the positively charged toner particles away from the drum.

3 0

3 years ago