A force of 7.50 N is applied to a spring whose spring constant is .298 N/cm. Find it’s length

A 50g ball is released from rest 1.0 above the bottom of thetrack

ludmilkaskok [199]

Answer:

The maximum height of the ball is 2 m.

Explanation:

Given that,

Mass of ball = 50 g

Height = 1.0 m

Angle = 30°

The equation is

$y=\dfrac{1}{4}x^2$

We need to calculate the velocity

Using conservation of energy

$\Delta U_{i}+\Delta K_{i}=\Delta K_{f}+\Delta U_{f}$

Here, ball at rest so initial kinetic energy is zero and at the bottom the potential energy is zero

$\Delta U_{i}=\Delta K_{f}$

Put the value into the formula

$mgh=\dfrac{1}{2}mv^2$

Put the value into the formula

$50\times10^{-3}\times9.8\times1.0=\dfrac{1}{2}\times50\times10^{-3}\times v^2$

$v^2=\dfrac{2\times50\times10^{-3}\times9.8\times1.0}{50\times10^{-3}}$

$v=\sqrt{19.6}$

$v=4.42\ m/s$

We need to calculate the maximum height of the ball

Using again conservation of energy

$\dfrac{1}{2}mv^2=mgh$

Here, h = y highest point

Put the value into the formula

$\dfrac{1}{2}\times50\times10^{-3}\times(4.42)^2=50\times10^{-3}\times9.8\times h$

$y=\dfrac{0.5\times(4.42)^2}{9.8}$

$y=0.996\ m$

Put the value of y in the given equation

$y=\dfrac{1}{4}x^2$

$x^2=4\times0.996$

$x=\sqrt{4\times0.996}$

$x=1.99\ m\ \approx 2 m$

Hence, The maximum height of the ball is 2 m.

4 0

3 years ago

Who submits the federal budget every year?

artcher [175]

C) The president submits the federal budget every year.
Hope this helps you!

8 0

3 years ago

A stone is thrown in a vertically

Lemur [1.5K]

Answer:

1.25 m

0.5 s

Explanation:

Given:

v₀ = 5 m/s

v = 0 m/s

a = -10 m/s²

Find: Δy

v² = v₀² + 2aΔy

(0 m/s)² = (5 m/s)² + 2 (-10 m/s²) Δy

Δy = 1.25 m

Find: t

v = at + v₀

(0 m/s) = (-10 m/s²) t + (5 m/s)

t = 0.5 s

6 0

4 years ago

A grapefruit falls from a tree and hits the ground .75 seconds later. How far did the grapefruit drop? What was its speed?

Ivanshal [37]

1) The grapefruit is in free fall, so it moves by uniformly accelerated motion, with constant acceleration $g=9.81 m/s^2$ . Calling h its height at t=0, the height at time t is given by
$h(t)=h- \frac{1}{2}gt^2$
We are told thatn when $t=0.75 s$ the grapefruit hits the ground, so h(0.75 s)=0. If we substitute these data into the equation, we can find the initial height h of the grapefruit:
$0=h- \frac{1}{2}gt^2$
$h= \frac{1}{2}gt^2= \frac{1}{2}(9.81 m/s^2)(0.75 s)^2=2.76 m$

2) The speed of the grapefruit at time t is given by
$v(t)=v_0 +gt$
where $v_0=0$ is the initial speed of the grapefruit. Substituting t=0.75 s, we find the speed when the grapefruit hits the ground:
$v(0.75 s)=gt=(9.81 m/s^2)(0.75 s)=7.36 m/s$

3 0

4 years ago

A uniform cylinder of radius 10 cm and mass 20 kg is mounted so as to rotate freely about a horizontal axis that is parallel to

fomenos

Answer:

a. 0.15 kg m2

b. 19.8 rad/s

Explanation:

Metric unit conversion:

10 cm = 0.1 m

5 cm = 0.05 m

a. Using parallel axis theorem, the rotational inertia of the cylinder about the axis of rotation is the inertia about the longitudinal axis plus the product of mass and distance from the longitudinal axis to the rotational axis squared

$I = I_L + md^2$

whereas the inertia about the longitudinal axis of the solid cylinder is

$I_L = mr^2/2 = 20*0.1^2/2 = 0.1 kgm^2$

$md^2 = 20*0.05^2 = 0.05 kgm^2$

$I = I_L + md^2 = 0.1 + 0.05 = 0.15 kg m^2$

b. Assume the cylinder does not rotate about its own longitudinal axis, we can treat this as a point mass pendulum. So when it's being released from 0.05m high (release point) to 0m (lowest position), its potential energy is converted to kinetic energy:

$E_p = E_k$

$mgh = mv^2/2$

where h = 0.05 is the vertical distance traveled, v is the cylinder linear velocity at the lowest position.g = 9.81m/s2 is the gravitational acceleration.

We can divide both sides by m

$gh = v^2/2$

$v^2 = 2gh = 2*9.81*0.05 = 0.981$

$v = \sqrt{0.981} = 0.99 m/s$

The angular speed is linear speed divided by the radius of rotation, which is distance from the cylinder center to the center of rotation d = 0.05 m

$\omega = \frac{v}{d} = \frac{0.99}{0.05} = 19.8 rad/s$

5 0

3 years ago