2. When pressing the accelerator on a

A ball is kicked at an angle of 35° with the ground.a) What should be the initial velocity of the ball so that it hits a target

stiks02 [169]

Answer:

a.18.5 m/s

b.1.98 s

Explanation:

We are given that

$\theta=35^{\circ}$

a.Let $v_0$ be the initial velocity of the ball.

Distance,x=30 m

Height,h=1.8 m

$v_x=v_0cos\theta=v_0cos35$

$v_y=v_0sin\theta=v_0sin35$

$x=v_0cos\theta\times t=v_0cos35\times t$

$t=\frac{30}{v_0cos35}$

$h=v_yt-\frac{1}{2}gt^2$

Substitute the values

$1.8=v_0sin35\frac{30}{v_0cos35}-\frac{1}{2}(9.8)(\frac{30}{v_0cso35})^2$

$1.8=30tan35-\frac{6574.6}{v^2_0}$

$\frac{6574.6}{v^2_0}=21-1.8=19.2$

$v^2_0=\frac{6574.6}{19.2}$

$v_0=\sqrt{\frac{6574.6}{19.2}}=18.5 m/s$

Initial velocity of the ball=18.5 m/s

b.Substitute the value then we get

$t=\frac{30}{18.5cos35}$

t=1.98 s

Hence, the time for the ball to reach the target=1.98 s

7 0

3 years ago

A 2500-ohm is connected to a 110v power supply. What is the current through the resistor

Assoli18 [71]

Answer:

0.044 amps

Explanation:

Givens

R = 2500 ohms

E = 110 volts

I = ?

Equation

I = E/R

Solution

I = 110/2500

I = 0.044 amps

7 0

3 years ago

Suppose a constant net force of 345 N is applied to slide a heavy stationary couch across the

ira [324]

Answer:

517.5Ns

Explanation:

F=(MV - MU)/t

where MV - MU is the change in momentum,

therefore, MV - MU = Ft

= 345 X 1.

= 517.5Ns

4 0

3 years ago

The 4kg head of a sledge hammer is moving at 6m/s when it strikes a chisel, driving it into a log. The duration of the impact (o

LUCKY_DIMON [66]

Answer:

The average impact force is 12000 newtons.

Explanation:

By Impact Theorem we know that impact done by the sledge hammer on the chisel is equal to the change in the linear momentum of the former. The mathematical model that represents the situation is now described:

$\bar F \cdot \Delta t = m \cdot (v_{2}-v_{1})$ (1)

Where:

$\bar F$ - Average impact force, in newtons.

$\Delta t$ - Duration of the impact, in seconds.

$m$ - Mass of the sledge hammer, in kilograms.

$v_{1}$ , $v_{2}$ - Initial and final velocity, in meters per second.

If we know that $\Delta t = 0.0020\,s$ , $m = 4\,kg$ , $v_{1} = -6\,\frac{m}{s}$ and $v_{2} = 0\,\frac{m}{s}$ , then we estimate the average impact force is:

$\bar F = \frac{m\cdot (v_{2}-v_{1})}{\Delta t}$

$\bar F = 12000\,N$

The average impact force is 12000 newtons.

5 0

3 years ago

A 5.50-kg object is hung from the bottom end of a vertical spring fastened to an overhead beam. The object is set into vertical

iris [78.8K]

Answer:

17.71N/m

Explanation:

The period of the spring is expressed according to the expression;

$T = 2 \pi \sqrt{\frac{m}{k} } \\$

m is the mass of the object

k is the force constant

Given

m = 5.50kg

T = 3.50s

Substitute into the formula;

$T = 2 \pi \sqrt{\frac{m}{k} } \\3.5 = 2 (3.14) \sqrt{\frac{5.5}{k} } \\3.5 = 6.28 \sqrt{\frac{5.5}{k} } \\\frac{3.5}{6.28} = \sqrt{\frac{5.5}{k} } \\0.557 = \sqrt{\frac{5.5}{k} } \\square \ both \ sides\\0.557^2 = (\sqrt{\frac{5.5}{k} })^2 \\0.3106 = \frac{5,5}{k}\\k = \frac{5.5}{0.3106}\\k = 17.71N/m$

Hence the force constant of the spring is 17.71N/m

4 0

3 years ago