All categories

1 year ago

FInal physics question please help asap

Physics

1 answer:

blondinia [14]1 year ago

5 0

The cannonball will be 311.8 m far in meter when it travels horizontally

<h3>What is a Projectile ?</h3>

A projectile is any object like stone, ball or javelin that take a parabolic path when thrown into the air.

The following parameters are given from the question

Ф = 30°

u = 60 m/s

g = 10 m/s²

R = ?

Resolve the initial velocity into horizontal and vertical components. That is,

Horizontal component = ucosФ

Vertical component = usinФ

To calculate the range that is, the horizontal distance, will can first calculate the total time take by the ball to reach the ground.

Total time T = 2usinФ / g

Substitute all the parameters

T = (2 x 60sin30) / 10

T = (2 x 30) / 10

T = 60 / 10

T = 6 s

The horizontal distance R = ucosФ x T

R = 60 x cos 30 x 6

R = 311.77 m

Therefore, the cannonball will travel 311.8 m approximately horizontally.

Learn more about Projectile here: brainly.com/question/24949996

#SPJ1

You might be interested in

Steam is to be condensed on the shell side of a heat exchanger at 150 oF. Cooling water enters the tubes at 60 oF at a rate of 4

zalisa [80]

Answer:

a. 572Btu/s

b.0.1483Btu/s.R

Explanation:

a.Assume a steady state operation, KE and PE are both neglected and fluids properties are constant.

From table A-3E, the specific heat of water is $c_p=1.0\ Btu/lbm.F$ , and the steam properties as, A-4E:

$h_{fg}=1007.8Btu/lbm, s_{fg}=1.6529Btu/lbm.R$

Using the energy balance for the system:

$\dot E_{in}-\dot E_{out}=\bigtriangleup \dot E_{sys}=0\\\\\dot E_{in}=\dot E_{out}\\\\\dot Q_{in}+\dot m_{cw}h_1=\dot m_{cw}h_2\\\\\dot Q_{in}=\dot m_{cw}c_p(T_{out}-T_{in})\\\\\dot Q_{in}=44\times 1.0\times (73-60)=572\ Btu/s$

Hence, the rate of heat transfer in the heat exchanger is 572Btu/s

b. Heat gained by the water is equal to the heat lost by the condensing steam.

-The rate of steam condensation is expressed as:

$\dot m_{steam}=\frac{\dot Q}{h_{fg}}\\\\\dot m_{steam}=\frac{572}{1007.8}=0.5676lbm/s$

Entropy generation in the heat exchanger could be defined using the entropy balance on the system:

$\dot S_{in}-\dot S_{out}+\dot S_{gen}=\bigtriangleup \dot S_{sys}\\\\\dot m_1s_1+\dot m_3s_3-\dot m_2s_2-\dot m_4s_4+\dot S_{gen}=0\\\\\dot m_ws_1+\dot m_ss_3-\dot m_ws_2-\dot m_ss_4+\dot S_{gen}=0\\\\\dot S_{gen}=\dot m_w(s_2-s_1)+\dot m_s(s_4-s_3)\\\\\dot S_{gen}=\dot m c_p \ In(\frac{T_2}{T_1})-\dot m_ss_{fg}\\\\\\\dot S_{gen}=4.4\times 1.0\times \ In( {73+460)/(60+460)}-0.5676\times 1.6529\\\\=0.1483\ Btu/s.R$

Hence,the rate of entropy generation in the heat exchanger. is 0.1483Btu/s.R

4 0

3 years ago

Two identical satellites orbit the earth in stable orbits. One satellite orbits with a speed v at a distance r from the center o

jolli1 [7]

Answer:

c)At a distance greater than r

Explanation:

For a satellite in orbit around the Earth, the gravitational force provides the centripetal force that keeps the satellite in motion:

$G\frac{Mm}{r^2}=m\frac{v^2}{r}$

where

G is the gravitational constant

M is the Earth's mass

m is the satellite's mass

r is the distance between the satellite and the Earth's centre

v is the speed of the satellite

Re-arranging the equation, we write

$r = \frac{GM}{v^2}$

so we see from the equation that when the speed is higher, the distance from the Earth's centre is smaller, and when the speed is lower, the distance from the Earth's centre is larger.

Here, the second satellite orbit the Earth at a speed less than v: this means that its orbit will have a larger radius than the first satellite, so the correct answer is

c)At a distance greater than r

7 0

3 years ago

A river flows due east at 1.60 m/s. A boat crosses the river from the south shore to the north shore by maintaining a constant v

Citrus2011 [14]

Answer:

part (a) $v\ =\ 10.42\ at\ 81.17^o$ towards north east direction.

part (b) s = 46.60 m

Explanation:

Given,

velocity of the river due to east = $v_r\ =\ 1.60\ m/s.$
velocity of the boat due to the north = $v_b\ =\ 10.3\ m/s.$

part (a)

River is flowing due to east and the boat is moving in the north, therefore both the velocities are perpendicular to each other and,

Hence the resultant velocity i,e, the velocity of the boat relative to the shore is in the North east direction. velocities are the vector quantities, Hence the resultant velocity is the vector addition of these two velocities and the angle between both the velocities are $90^o$

Let 'v' be the velocity of the boat relative to the shore.

$\therefore v\ =\ \sqrt{v_r^2\ +\ v_b^2}\\\Rightarrow v\ =\ \sqrt{1.60^2\ +\ 10.3^2}\\\Rightarrow v\ =\ 10.42\ m/s.$

Let $\theta$ be the angle of the velocity of the boat relative to the shore with the horizontal axis.

Direction of the velocity of the boat relative to the shore. $\therefore Tan\theta\ =\ \dfrac{v_b}{v_r}\\\Rightarrow Tan\theta\ =\ \dfrac{10.3}{1.60}\\\Rightarrow \theta\ =\ Tan^{-1}\left (\dfrac{10.3}{1.60}\ \right )\\\Rightarrow \theta\ =\ 81.17^o$

part (b)

Width of the shore = w = 300m

total distance traveled in the north direction by the boat is equal to the product of the velocity of the boat in north direction and total time taken

Let 't' be the total time taken by the boat to cross the width of the river. $\therefore w\ =\ v_bt\\\Rightarrow t\ =\ \dfrac{w}{v_b}\\\Rightarrow t\ =\ \dfrac{300}{10.3}\\\Rightarrow t\ =\ 29.12 s$

Therefore the total distance traveled in the direction of downstream by the boat is equal to the product of the total time taken and the velocity of the river $\therefore s\ =\ u_rt\\\Rightarrow s\ =\ 1.60\times 29.12\\\Rightarrow s\ =\ 46.60\ m$

7 0

3 years ago

Net force of 8.0 N acts on an 18 kg body for one minute. Determine the impulse due to the force.

boyakko [2]

Answer:

p = FΔt = 8.0 N(60 s) = 480 N•s

Explanation:

not asked for, but in that time a frictionless 18 kg mass on a horizontal surface will have change velocity by 480/18 = 26.7 m/s.

An impulse results in a change of momentum.

3 0

3 years ago

What is the relationship between mass gravity and weight?

Basile [38]

Fg=m•g || IE: Weight = mass x gravity
Therefore, the relationship are as follows:
mass and gravity are inversely proportional
mass and weight are directly proportional
weight and gravity are directly proportional

7 0

3 years ago