Could someone please help me

A hockey puck slides off the edge of a table with an initial velocity of 23.2 m/s and experiences no air resistance. The height

Dennis_Churaev [7]

Answer:

15.1°

Explanation:

The horizontal velocity of the hockey puck is constant during the motion, since there are no forces acting along this direction:

$v_x = 23.2 m/s$

Instead, the vertical velocity changes, due to the presence of the acceleration due to gravity:

$v_y(t)= v_{y0} -gt$ (1)

where

$v_{y0}=0$ is the initial vertical velocity

g = 9.8 m/s^2 is the gravitational acceleration

t is the time

Since the hockey puck falls from a height of h=2.00 m, the time it needs to reach the ground is given by

$h=\frac{1}{2}gt^2\\t=\sqrt{\frac{2h}{g}}=\sqrt{\frac{2(2.00 m)}{9.8 m/s^2}}=0.64 s$

Substituting t into (1) we find the final vertical velocity

$v_y = -(9.8 m/s^2)(0.64 s)=-6.3 m/s$

where the negative sign means that the velocity is downward.

Now that we have both components of the velocity, we can calculate the angle with respect to the horizontal:

$tan \theta = \frac{|v_y|}{v_x}=\frac{6.3 m/s}{23.2 m/s}=0.272\\\theta = tan^{-1} (0.272)=15.1^{\circ}$

6 0

3 years ago

You're in a car that gets 28 miles per gallon of gas, driving it at a constant speed. If you took the gas from the car tank, and

Nataly [62]

Answer:

The diameter of the hose is 0.326 mm.

Explanation:

Given that,

Speed of car = 28 miles/galllon

We need to calculate the radius

The rate of flow of fluid is from the equation of continuity

$\dfrac{V}{t}=Av$

Where, A = area of cross sectional

$\dfrac{V}{t}=\pi r^2v$

We know that,

The velocity is the ratio of displacement of gas per unit time.

$\dfrac{V}{t}=\pi r^\dfrac{x}{t}$

$\dfrac{V}{x}=\pi r^2$

Put the value into the formula

$\dfrac{0.00378541}{28\times1609.34}=\pi r^2$

$r=\sqrt{\dfrac{0.00378541}{28\times1609.34\times\pi}}$

$r=1.63\times10^{-4}\ m$

We need to calculate the diameter of the hose

$d = 2r$

Put the value of r

$d=2\times1.63\times10^{-4}$

$d=0.326\ mm$

Hence, The diameter of the hose is 0.326 mm.

6 0

3 years ago

What is the energy stored between 2 Carbon nuclei that are 1.00 nm apart from each other? HINT: Carbon nuclei have 6 protons and

Andrei [34K]

Answer:

$A. 8.29\times 10^{-18}\ J$

Explanation:

Given that:

p = magnitude of charge on a proton = $1.6\times 10^{-19}\ C$

k = Boltzmann constant = $9\times 10^{9}\ Nm^2/C^2$

r = distance between the two carbon nuclei = 1.00 nm = $1.00\times 10^{-9}\ m$

Since a carbon nucleus contains 6 protons.

So, charge on a carbon nucleus is $q = 6p=6\times 1.6\times 10^{-19}\ C=9.6\times 10^{-19}\ C$

We know that the electric potential energy between two charges q and Q separated by a distance r is given by:

$U = \dfrac{kQq}{r}$

So, the potential energy between the two nuclei of carbon is as below:

$U= \dfrac{kqq}{r}\\\Rightarrow U = \dfrac{kq^2}{r}\\\Rightarrow U = \dfrac{9\times10^9\times (9.6\times 10^{-19})^2}{1.0\times 10^{-9}}\\\Rightarrow U =8.29\times 10^{-18}\ J$

Hence, the energy stored between two nuclei of carbon is $8.29\times 10^{-18}\ J$ .

8 0

3 years ago

g If you keep the launch angle fixed, but double the initial launch speed, what happens to the range?

kiruha [24]

Answer:

Range will become 4 times of initial range

Explanation:

Let the velocity of projection is u

And angle at which projectile is projected is $\Theta$

And acceleration due to gravity is $g\ m/sec^2$

So range of projectile is equal to $R=\frac{u^2sin2\Theta }{g}$ ........eqn 1

Now in second case it is given that velocity of launching is doubled

So new velocity $u_{new}=2u$

So new range will be equal to $R_{new}=\frac{(2u)^2sin2\Theta }{g}=\frac{4u^2sin2\Theta }{g}$ .....eqn 2

Now dividing eqn 2 by eqn 1

$\frac{R_{new}}{R}=\frac{4u^2sin2\Theta }{g}\times \frac{g}{u^2sin2\Theta }$

$R_{new}=4R$

So if we double the initial launch speed then range will become 4 times

4 0

4 years ago

Under electrostatic conditions, the electric field just outside the surface of any charged conductor

KatRina [158]

Answer:

C. is always perpendicular to the surface of the conductor

Explanation:

On a charged conductor , electric charge is uniformly distributed on its surface . The lines of forces are also uniformly distributed on all directions . They repel each other so they emerge perpendicular to the surface so that they do nor cut each other and at the same time they remain at maximum distance from each other.

3 0

3 years ago