PHYSICS COLLEGE

Could you please solve it with shiwing the full work1-

A high powered projectile is fired horizontally from the top of a cliff at a speed of 638.6 m/s. Determine the magnitude of the velocity (in m/s) after 5 seconds.

Take gravitational acceleration to be 9.81 m/s2.

2-

A man throws a ball with a velocity of 20.9 m/s upwards at 33.2° to the horizontal. At what vertical distance above the release height (in metres) will the ball strike a wall 13.0 m away ?

Take gravitational acceleration to be 9.81 m/s2.

3-

A particle is moving along a straight path and its position is defined by the equation s = (1t3 + -5t2 + 3) m, where t is measured in seconds. Determine the average velocity (in m/s) of the particle when t = 5 seconds.

4-

A particle has an initial speed of 26 m/s. The particle undergoes a deceleration of a = (-9t) m/s2, where t is measured in seconds. Determine the distance (in metres) the particle travels before it stops. When t = 0, s = 0.

Answers

Answer 1

Answer:

Answer:

1.V= 640.48 m/s :total velocity in t= 5s

2. Y= 5.79m : vertical distance above the height of release (in meters) where the ball will hit a wall 13.0 m away

3. v =25m/s

4. s= (-1.5t³+26t ) m

Explanation:

1. Parabolic movement in the x-y plane , t=5s

V₀=638.6 m/s=Vx :Constant velocity in x

Vy=V₀y +gt= 0+9.8*5 = 49 m/s : variable velocity in y

$v=\sqrt{v_(x) ^(2) +v_(y) ^(2) }$

$v=\sqrt{ 638.6^(2) +49 ^(2) }$

V= 640.48 m/s : total velocity in t= 5s

2. $v_(ox) =v_(o) cos33.2=20.9*cos33,2= 17.49 m/s$

$v_(oy)=v_(o)*sin33,2 =20.9*sin33,2=11.44 m/s$

x=v₀x*t

13=v₀x*t

13=17.49*t

t=13/17.49=0.743s : time for 13.0 m away

th=v₀y/g=11.44/9.8= 1,17s :time for maximum height

at t=0.743 sthe ball is going up ,then g is negative

y=v₀y*t - 1/2 *g¨*t²

y=11.44*0.743 -1/2*9.8*0.743²

y= 5.79m : vertical distance above the height of release (in meters) where the ball will hit a wall 13.0 m away

3. s = (1t3 + -5t2 + 3) m

v=3t²-10t=3*25-50=75-50=25m/s

at t=0, s=3 m

at t=5s s=5³-5*5²+3

4. a = (-9t) m/s2

a=dv/dt=-9t

dv=-9tdt

v=∫ -9tdt

v=-9t²/2 + C1 equation (1)

in t=0 , v₀=26m/s ,in the equation (1) C1= 26

v=-9t²/2 + 26=ds/dt

ds=( -9t²/2 + 26)dt

s= ∫( -9t²/2 + 26)dt

s= -9t³/6+26t+C2 Equation 2

t = 0, s = 0 , C2=0

s= (-9t³/6+26t ) m

s= (-1.5t³+26t ) m

Related Questions

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What is the average kinetic energy of hydrogen atoms on the 5500°C surface of the sun?
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Can a car moving with a negative velocity moves faster than a car moving with a positive velocity? explain.

An object is made from a material that is hard, shiny, and can be hammered into thin sheets. Which of these is another likely property of the material?

Answers

Answer:

Good conductor of heat

Explanation:

Because metals are shiny, ductile, malleable, sonorous, good conductors of heat and electricity and have high melting and boiling points

Answer: mine is different so im sorry im here for points

Explanation:

A 0.47 kg block of wood hangs from the ceiling by a string, and a 0.070kg wad of putty is thrown straight upward, striking the bottom of the block with a speed of 5.60 m/s. The wad of putty sticks to the block. (Answer on previous exams) How high does the putty-block system rise above the original position of the block Is the kinetic energy of the system conserved during the collision Is the mechanical energy of the system conserved during the collision Is the mechanical energy conserved after the collision

Answers

Answer:

The height is $h =0.0269 \ m$

The kinetic energy during collision is not conserved

The Mechanical energy during the collision is not conserved

The mechanical energy after the collision is not conserved

Explanation:

From the question we are told that

The mass of the block is $m_b = 0.47\ kg$

The mass of the wad of putty is $m_p = 0.070 \ kg$

The speed o the wad of putty is $v_p = 5.60 \ m/s$

The law of momentum conservation can be mathematically represented as

$p_i = p_f$

Where $p_i$ is the initial momentum which is mathematically represented as

$p_i =m_p * v_p$

While $p_f$ is the initial momentum which is mathematically represented as

$p_f = (m_b + m_p)v_f$

Where $v_f$ s the final velocity

$m_p v_p = (m_p + m_b) * v_f$

Making $v_f$ the subject

$v_f = (m_p v_p)/(m_b +m_p)$

substituting values

$v_f = ((0.070)*(5.60))/(0.47 + 0.070)$

$v_f = 0.726 \ m/s$

According to the law of energy conservation

$KE = PE$

Where KE is the kinetic energy of the system which is mathematically represented as

$KE = (1)/(2) (m_p + m_b)v_f^2$

And PE is the potential energy of the system which is mathematically represented as

$PE = (m_p +m_b) gh$

$(1)/(2) (m_p + m_b)v_f^2 = (m_p +m_b) gh$

Making h the subject of the formula

$h = (v_f^2)/(2g)$

substituting values

$h = ((0.726 )^2 )/(2 * 9.8)$

$h =0.0269 \ m$

Now the kinetic energy is conserved during collision because the system change it height during which implies some of the kinetic energy was converted to potential energy during collision

The the mechanical energy of the system during the collision is conserved because this energy consists of the kinetic and the potential energy.

Now after the collision the mechanical energy is not conserved because the external force like air resistance has reduced the mechanical energy of that system

A 2100 g block is pushed by an external force against a spring (with a 22 N/cm spring constant) until the spring is compressed by 11 cm from its uncompressed length. The compressed spring and block rests at the bottom of an incline of 28◦ with the spring lying along the surface of the ramp.After all the external forces are removed (so the compressed spring releases the mass) how far D along the plane will the block move before coming to a stop? Answer in units of m.

Answers

Answer:

6.5e-4 m

Explanation:

We need to solve this question using law of conservation of energy

Energy at the bottom of the incline= energy at the point where the block will stop

Therefore, Energy at the bottom of the incline consists of the potential energy stored in spring and gravitational potential energy= $(1)/(2) kx^(2) +PE1$

Energy at the point where the block will stop consists of only gravitational potential energy= $PE2$

Hence from Energy at the bottom of the incline= energy at the point where the block will stop

⇒ $(1)/(2) kx^(2) +PE1=PE2$

⇒ $PE2-PE1=(1)/(2) kx^(2)$

Also $PE2-PE2=mgh$

where $m$ is the mass of block

$g$ is acceleration due to gravity=9.8 m/s

$h$ is the difference in height between two positions

⇒ $mgh=(1)/(2) kx^(2)$

Given m=2100kg

k=22N/cm=2200N/m

x=11cm=0.11 m

∴ $2100*9.8*h=(1)/(2)*2200*0.11^(2)$

⇒ $20580*h=13.31$

⇒ $h=(13.31)/(20580)$

⇒h=0.0006467m= $6.5e-4$

The fundamental force that is responsible for beta decay is the: A) weak force. B) strong force. C) dark energy force. D) gravitational force E) electromagnetic force.

Answers

Answer:

A. Weak forces

Explanation:

The fundamental forces responsible for beta decay is the weak force. Weak forces are among the four fundamental forces of universe the electromagnetic, gravitational and strong forces. The weak forces are responsible for the decaying. The fundamental work of weak forces is covert neutron into proton and electron into neutrino. weak forces operate at very low distances as low as fermi meter.

Answer:

The answer is

dark energy force.

hope this helps u stay safe

Explanation:

A 0.454-kg block is attached to a horizontal spring that is at its equilibrium length, and whose force constant is 21.0 N/m. The block rests on a frictionless surface. A 5.30×10?2-kg wad of putty is thrown horizontally at the block, hitting it with a speed of 8.97 m/s and sticking.Part AHow far does the putty-block system compress the spring?

Answers

The distance the putty-block system compress the spring is 0.15 meter.

Given the following data:

Mass = 0.454 kg

Spring constant = 21.0 N/m.

Mass of putty = $5.30* 10^(-2)\;kg$

Speed = 8.97 m/s

To determine how far (distance) the putty-block system compress the spring:

First of all, we would solver for the initialmomentum of the putty.

$P_p = mass * velocity\n\nP_p = 5.30* 10^(-2)* 8.97\n\nP_p = 47.54 * 10^(-2) \;kgm/s$

Next, we would apply the law of conservation of momentum to find the final velocity of the putty-block system:

$P_p = (M_b + M_p)V\n\n47.54* 10^(-2) = (0.454 + 5.30* 10^(-2))V\n\n47.54* 10^(-2) = 0.507V\n\nV = (0.4754)/(0.507)$

Velocity, V = 0.94 m/s

To find the compression distance, we would apply the law of conservation of energy:

$U_E = K_E\n\n(1)/(2) kx^2 = (1)/(2) mv^2\n\nkx^2 =M_(bp)v^2\n\nx^2 = (M_(bp)v^2)/(k) \n\nx^2 = ((0.454 + 5.30* 10^(-2)) * 0.94^2)/(21)\n\nx^2 = ((0.507 * 0.8836))/(21)\n\nx^2 = ((0.4480))/(21)\n\nx=√(0.0213)$

x = 0.15 meter

Answer:

Explanation:

Force constant of spring K = 21 N /m

we shall find the common velocity of putty-block system from law of conservation of momentum .

Initial momentum of putty

= 5.3 x 10⁻² x 8.97

= 47.54 x 10⁻² kg m/s

If common velocity after collision be V

47.54 x 10⁻² = ( 5.3x 10⁻² + .454) x V

V = .937 m/s

If x be compression on hitting the putty

1/2 k x² = 1/2 m V²

21 x² = ( 5.3x 10⁻² + .454) x .937²

x² = .0212

x = .1456 m

14.56 cm

An electron is accelerated by a 5.9 kV potential difference. das (sd38882) – Homework #9 – yu – (44120) 3 The charge on an electron is 1.60218 × 10−19 C and its mass is 9.10939 × 10−31 kg. How strong a magnetic field must be experienced by the electron if its path is a circle of

Answers

Complete Question

Answer:

The magnetic field strength is $B= 0.0048 T$

Explanation:

The work done by the potential difference on the electron is related to the kinetic energy of the electron by this mathematical expression

$\Delta V q = (1)/(2)mv^2$

Making v the subject

$v = \sqrt{[(2 \Delta V * q )/(m)] }$

Where m is the mass of electron

v is the velocity of electron

q charge on electron

$\Delta V$ is the potential difference

Substituting values

$v = \sqrt{(2 * 5.9 *10^3 * 1.60218*10^(-19) )/(9.10939 *10^(-31]) )$ f

$= 4.5556 *10^ {7} m/s$

For the electron to move in a circular path the magnetic force[ $F = B q v$ ] must be equal to the centripetal force[ $(mv^2)/(r)$ ] and this is mathematically represented as