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3 years ago

A ball is thrown into the air at some angle. At the very top of the ball's path, its velocity is

1 answer:

7 0

Entirely horizontal.

An object will continue to travel with the same velocity unless an external force is applied.
There are no external forces acting in the horizontal direction (except air resistance but the is negligable) so the ball will move in the horizonontal direction with with same speed at all times.
Force of gravity is acting downwards, meaning it is constantly acelleraring at 9.8ms^-2 towards the ground. When the ball reaches its heighest point, it is where the ball has stopped moving up and starts to move down, therfore it is not moving vertically at all.

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Homes with multiple bathrooms have outlets connected in series with a red reset button in the system

dybincka [34]

Homes with multiple bathrooms have outlets connected in series with a red reset button in the system. The reason for this circuit design is to prevent electrocution.

<h3>What is red reset button?</h3>

The water heater's reset switch provided on the upper thermostat of an electric water heater. If the button is on, the switch would have tripped and the reset is required.

To prevent any kind of injury or killing of a person by electric shock is called as the electrocution. This can be prevented only by outlets connected in series with a red reset button.

Thus, the reason for this circuit design is to prevent electrocution.

Learn more about red reset button.

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8 0

2 years ago

Write your question here (Keep it simple and clear to get the best answer) a car accelerates uniformly from an initial velocity

Fittoniya [83]

Answer:

Average velocity = 18 m/s

Explanation:

Given the following data;

Initial velocity = 10m/s

Acceleration = 2m/s²

Time = 4 seconds

To find the average velocity, we would use the first equation of motion;

$V = U + at$

Where;

V is the final velocity.

U is the initial velocity.

a is the acceleration.

t is the time measured in seconds.

Substituting into the equation, we have;

V = u + at

V = 10 + 2*4

V = 10 + 8

V = 18 m/s

8 0

2 years ago

A coaxial cable has a charged inner conductor (with charge +8.5 µC and radius 1.304 mm) and a surrounding oppositely charged con

Tcecarenko [31]

Complete question:

A 50 m length of coaxial cable has a charged inner conductor (with charge +8.5 µC and radius 1.304 mm) and a surrounding oppositely charged conductor (with charge −8.5 µC and radius 9.249 mm).

Required:

What is the magnitude of the electric field halfway between the two cylindrical conductors? The Coulomb constant is 8.98755 × 10^9 N.m^2 . Assume the region between the conductors is air, and neglect end effects. Answer in units of V/m.

Answer:

The magnitude of the electric field halfway between the two cylindrical conductors is 5.793 x 10⁵ V/m

Explanation:

Given;

charge of the coaxial capable, Q = 8.5 µC = 8.5 x 10⁻⁶ C

length of the conductor, L = 50 m

inner radius, r₁ = 1.304 mm

outer radius, r₂ = 9.249 mm

The magnitude of the electric field halfway between the two cylindrical conductors is given by;

$E = \frac{\lambda}{2\pi \epsilon_o r} = \frac{Q}{2\pi \epsilon_o r L}$

Where;

λ is linear charge density or charge per unit length

r is the distance halfway between the two cylindrical conductors

$r = r_1 + \frac{1}{2}(r_2-r_1) \\\\r = 1.304 \ mm \ + \ \frac{1}{2}(9.249 \ mm-1.304 \ mm)\\\\r = 1.304 \ mm \ + \ 3.9725 \ mm\\\\r = 5.2765 \ mm$

The magnitude of the electric field is now given as;

$E = \frac{8.5*10^{-6}}{2\pi(8.85*10^{-12})(5.2765*10^{-3})(50)} \\\\E = 5.793*10^5 \ V/m$

Therefore, the magnitude of the electric field halfway between the two cylindrical conductors is 5.793 x 10⁵ V/m

5 0

3 years ago

A force of 385 N is applied in pushing a stalled automobile at a constant speed for a distance of 150 m. How much work (in J) wa

Helga [31]

Answer:57,750J

Explanation:

3 0

2 years ago

A bullet of mass 11 g strikes a ballistic pendulum of mass 1.9 kg. The center of mass of the pendulum rises a vertical distance

Ymorist [56]

Answer:

The bullet's initial speed is 243.21 m/s.

Explanation:

Given that,

Mass of the bullet, $m_b=11\ g=0.011\ kg$

Mass of the pendulum, $m_p=19\ kg$

The center of mass of the pendulum rises a vertical distance of 10 cm.

We need to find the bullet's initial speed if it is assumed that the bullet remains embedded in the pendulum. Let it is v. In this case, the energy of the system remains conserved. The kinetic energy of the bullet gets converted to potential energy for the whole system. So,

$\dfrac{1}{2}(m_b+m_p)V^2 =(m_b+m_p)gh\\\\V=\sqrt{2gh} \ .................(1)$

V is the speed of the bullet and pendulum at the time of collision

Now using conservation of momentum as :

$m_bv=(m_p+m_b)V$

Put the value of V from equation (1) in above equation as :

$v=\dfrac{(m_p+m_b)}{m_b}\sqrt{2gh} \\\\v=\dfrac{(1.9+0.011)}{0.011}\sqrt{2\times 9.8\times 0.1}\\\\v=243.21\ m/s$

So, the bullet's initial speed is 243.21 m/s.

7 0

3 years ago