Two astronauts on opposite ends of a spaceship are comparing lunches. One has an apple, the other has an orange. They decide to trade. Astronaut 1 tosses the 0.110 kg apple toward astronaut 2 with a speed of vi,1 = 1.13 m/s . The 0.150 kg orange is tossed from astronaut 2 to astronaut 1 with a speed of 1.25 m/s . Unfortunately, the fruits collide, sending the orange off with a speed of 0.977 m/s in the negative y direction.

Answers

Answer 1

Answer:

the final velocity of the apple after the collision is approximately 0.758m/s in the positive x-direction.

To solve this problem, we can use the principle of conservation of linear momentum. The total momentum before the collision should equal the total momentum after the collision.

Let's set up our coordinate system with the x-axis pointing to the right and the y-axis pointing upward. Astronaut 1 is tossing the apple in the positive x-direction, so the velocity of the apple (v1) will be positive. Astronaut 2 is tossing the orange in the negative x-direction, so the velocity of the orange (v2) will be negative.

The conservation of linear momentum equation is as follows:

m1∗v1+m2∗v2=m1∗vf1+m2∗vf2

Where:

m1 is the mass of the apple (0.110 kg)

v1 is the initial velocity of the apple (1.13 m/s)

m2 is the mass of the orange (0.150 kg)

v2 is the initial velocity of the orange (−1.25 m/s, as it's in the negative x-direction)

vf1 is the final velocity of the apple (which we need to find)

vf2 is the final velocity of the orange (−0.977 m/s)

Now, we can plug in these values and solve for vf1:

0.110kg∗1.13m/s+0.150kg∗(−1.25m/s)=0.110kg∗vf1+0.150kg∗(−0.977m/s)

0.1243kg∗m/s−0.1875kg∗m/s=0.110kg∗vf1−0.14655kg∗m/s

Now, let's isolate vf1:

0.1243kg∗m/s−0.1875kg∗m/s+0.14655kg∗m/s=0.110kg∗vf1

0.0834kg∗m/s=0.110kg∗vf1

Now, divide by 0.110kg to find vf1:

vf1=0.0834kg∗m/s/0.110kg=0.758m/s

So, the final velocity of the apple after the collision is approximately 0.758m/s in the positive x-direction.

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Answer 2

Answer:

Answer:

The speed and direction of the apple is 1.448 m/s and 66.65°.

Explanation:

Given that,

Mass of apple = 0.110 kg

Speed = 1.13 m/s

Mass of orange = 0.150 kg

Speed = 1.25 m/s

Suppose we find the final speed and direction of the apple in this case

Using conservation of momentum:

Before:

In x direction,

$P_(b)=m_(p)v_(p)-m_(o)v_(o)$

$P_(b)=0.110*1.13-0.150*1.25$

$P_(b)=−0.0632\ kg-m/s$

In y direction = 0

After:

$v_(ay)$ is velocity of the apple in the y direction

$v_(ax)$ is the velocity of the apple in the x direction

Momentum again:

In x direction,

$0.110* v_(ax)+0=−0.0632$

$v_(x)=(−0.0632)/(0.110)$

$v_(x)=−0.574\ m/s$

In y-direction,

$0.110* v_(ay)-0.150*0.977=0$

$v_(ay)=(0.150*0.977)/(0.110)$

$v_(ay)=1.33\ m/s$

We need to calculate the speed of apple

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

Put the value into the formula

$v_(a)=√((−0.574)^2+(1.33)^2)$

$v_(a)=1.448\ m/s$

We need to calculate the direction of the apple

Using formula of angle

$\tan\theta=(v_(ay))/(v_(ax))$

Put the value into the formula

$\theta=\tan^(-1)((1.33)/(0.574))$

$\theta=66.65^(\circ)$

Hence, The speed and direction of the apple is 1.448 m/s and 66.65°.

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Express the following speeds as a function of the speed of light, c: (a) an automobile speed (93 km/h) (b) the speed of sound (329 m/s) (c) the escape velocity of a rocket from the Earth's surface (12.1 km/s) (d) the orbital speed of the Earth about the Sun (Sun-Earth distance 1.5×108 km).

Answers

Answer:

(a). An automobile speed as a function of speed of light is $8.61*10^(-8)\ m/s$

(b). The speed of sound as a function of speed of light is $10.9*10^(-8)\ m/s$

(c). The escape velocity of a rocket as a function of speed of light is $4.03*10^(-8)\ m/s$

(d). The orbital speed of the Earth about the Sun is $3*10^(8)\ m/s$

Explanation:

Given that,

Express the following speeds as a function of the speed of light.

Automobile speed = 93 km/h

We know that,

A function of speed of light is

$c=3*10^(8)\ m/s$

(a). Automobile speed = 93 km/h

Speed $v_(a)=93*(5)/(18)$

$v_(a)=25.83\ m/s$

We need to express the speed of automobile speed as a function of speed of light

Using formula of speed

$v=(v_(a))/(v_(l))$

Put the value into the formula

$v=(25.83)/(3*10^(8))$

$v=8.61*10^(-8)\ m/s$

(b). The speed of sound is 329 m/s.

We need to express the speed of sound as a function of speed of light

Using formula of speed

$v=(v_(s))/(v_(l))$

Put the value into the formula

$v=(329)/(3*10^(8))$

$v=10.9*10^(-8)\ m/s$

(c). The escape velocity of a rocket from the Earth's surface is 12.1 m/s

We need to express the speed of sound as a function of speed of light

Using formula of speed

$v=(v_(e))/(v_(l))$

Put the value into the formula

$v=(12.1)/(3*10^(8))$

$v=4.03*10^(-8)\ m/s$

(d). The orbital speed of the Earth about the Sun

Distance = 1.5\times10^{8}[/tex]

We know that,

The sun rays reached on the earth in 8 min 20 sec.

We need to express the speed of sound as a function of speed of light

Using formula of speed

$v=(d)/(t)$

Put the value into the formula

$v=(1.5*10^(8)*1000)/(500)$

$v=3*10^(8)\ m/s$

Hence, (a). An automobile speed as a function of speed of light is $8.61*10^(-8)\ m/s$

(b). The speed of sound as a function of speed of light is $10.9*10^(-8)\ m/s$

(c). The escape velocity of a rocket as a function of speed of light is $4.03*10^(-8)\ m/s$

(d). The orbital speed of the Earth about the Sun is $3*10^(8)\ m/s$

A vibrating tuning fork makes 500 vibrations in one second. What is the wavelength of the sound produced if the air temperature is 20°C? Group of answer choices0.500 m
0.686 m
0.343 m
1.46 m
1.87 m

Answers

Answer:

wavelength will be 0.686 m

So option (b) is correct

Explanation:

We have given vibration in one second that is frequency = 500 Hz

We know that velocity of sound is 343 m/sec

We have to find the wavelength

We know that velocity is given by $v=\lambda f$

So wavelength $\lambda =(v)/(f)=(343)/(500)=0.686m$

So wavelength will be 0.686 m

So option (b) is correct

A CO2 gun shoots a 0.2 gram round pellet (bb) at 2800 ft/sec, and as the bb leaves the gun it gets charged by friction . If Earths magnetic field points South to North at an intensity of 20 uT, and the bb is shot W->E. Find the charge the bb would need to stay level by balancing out the force of gravity.

Answers

Answer:

So it would need a charge of 0.115C for a upward force to act and cancel the force of gravity.

Explanation:

Fb = Fg

qvb= mg ⇒ q = mg/vB = 0.2 *10∧-3 * 9.8/853.44 * 20 * 10∧-6

= 0.115C

note:2800ft/sec = 853.44m/s

So it would need a charge of 0.115C for a upward force to act and cancel the force of gravity.

Q.1- Find the distance travelled by a particle moving in a straight line with uniform acceleration, in the 10th unit of time.

Answers

Answer:

If the acceleration is constant, the movements equations are:

a(t) = A.

for the velocity we can integrate over time:

v(t) = A*t + v0

where v0 is a constant of integration (the initial velocity), for the distance traveled between t = 0 units and t = 10 units, we can solve the integral:

$\int\limits^(10)_0 {A*t + v0} \, dt = ((A/2)10^2 + v0*10) = (A*50 + v0*10)$

Where to obtain the actual distance you can replace the constant acceleration A and the initial velocity v0.

A physics cart has a projectile launcher mounted on top. While traveling on a straight track at 0.500 m/s, a projectile is fired. It lands back in the same place on top of the launcher after the cart has moved a distance of 2.30 m. In the frame of reference of the cart, (a) at what angle was the projectile fired and (b) what was the initial velocity of the projectile? (c) What is the shape of the projectile as seen by an observer on the cart? A physics student is watching the demonstration from a classroom seat. According to the student, (d) what is the shape of the projectile’s path, and (e) what is its initial velocity?

Answers

Answer:

(a) $90^(\circ)$

(b) Initial velocity of the projectile is 22.54 m/s

(d) Parabolic

(e) The initial velocity is 23.04 m/s

Solution:

As per the question:

Velocity of the cart, v = 0.500 m/s

Distance moved by the cart, d = 2.30 m

Now,

(a) The projectile must be fired at an angle of $90^(\circ)$ so that it mounts on the top of the cart moving with constant velocity.

(b) Now, for initial velocity, u':

Time of flight is given by;

$T = (D)/(v)$ (1)

where

T = Flight time

D = Distance covered

(b) The component of velocity w.r.t an observer:

Horizontal component, $v_(x) = u'cos\theta$

Vertical component, $v_(y) = u'sin\theta - gT$

Also, the vertical component of velocity at maximum height is zero, $v_(y) = 0$

Therefore, $T = (u')/(g)$

Total flight time, $(2u')/(g)$ (2)

Now, from eqn (1) and (2):

$u' = (gD)/(2v)$

$u' = (9.8* 2.30)/(2* 0.500) = 22.54 m/s$

(d) The shape of the path of the projectile seen by the physics student outside the reference frame of the cart is parabolic

(e) The initial velocity is given by:

u = u' + v = 22.54 + 0.5 = 23.04 m/s

A small ball of mass m is held directly above a large ball of mass M with a small space between them, and the two balls are dropped simultaneously from height H. (The height is much larger than the radius of each ball, so you may neglect the radius.) The large ball bounces elastically off the floor and the small ball bounces elastically off the large ball. a) For which value of the mass m, in terms of M, does the large ball stop when it collides with the small ball? b) What final height, in terms of H, does the small ball reach?

Answers

a) The large sphere has 3 times the mass of the small sphere

b) The final height at which small ball reach y = 4H

What will be the mass of the sphere and height covered by the small ball?

We must start this problem by calculating the speed with which the spheres reach the floor

$V_f^2=V_o^2-2gy$

As the spheres are released v₀ = 0

$V_f^2=2gH$

$V_f=√(2gH)$

The two spheres arrive at the same speed to the floor.

The largest sphere clashes elastically so that with the floor it has a much higher mass, the sphere bounces with the same speed with which it arrived, the exit speed of the spheres

$V_(10)=\sqrt2gH$

The big sphere goes up and the small one down, the two collide, let's form a system that contains the two spheres, let's use moment conservation

Let's call

$V_h=\sqrt2gH$

Small sphere m₂ and $V_(20)=-\sqrt2gH=-V_h$

Large sphere m₁ and $V_(10)=√(2gH)=V_h$

Before crash

$P_o=m_1V_(10)+m_2V_(20)$

After the crash

$P_f=m_1V_(1f)+m_2V_(2f)$

$P_o=P_f$

$m_1V_(10)+m_2V_(20)=m_1V_(1f)+m_2V_(2f)$

The conservation of kinetic energy

$K_o=(1)/(2) m_1V_(10)^2+(1)/(2) m_2V_(20)^2$

$K_f=(1)/(2) m_1V_(1f^2)+(1)/(2) m_2V_(2f)^2$

$K_o=K_f$

$K_o=(1)/(2) m_1V_(10)^2+(1)/(2) m_2V_(20)^2$ $=$ $K_f=(1)/(2) m_1V_(1f^2)+(1)/(2) m_2V_(2f)^2$

Let's write the values

$-m_1V_h+m_2V_h=m_1V_(1f)+m_2V_(2f)$

$m_1V_h^2+m_2V_h^2=m_1V_(1f)^2+m_2V_(2f)^2$

The solution to this system of equations is

$m_t=m_1+m_2$

$V_(1f)=((m_1-m_2))/(m_tV_(10)) +(2m_2)/(m_tV_2)$

$V_(2f)=(2m_1)/(m_tV_(10)) +(m_2-m_1)/(m_tV_2)$

The large sphere is labeled 1, we are asked for the mass so that $V_(1f)$ = 0, let's clear the equation

$V_(1f)=(m_1-m_2)/(m_tV_(10)) +(2m_2)/(m_tV_(20))$

$0=(m_1-m_2)/(m_tV_h) +(2m_2)/(m_t(-V_h))$

$(m_1-m_2)/(m_tV_h) =(2m_2)/(m_tV_h)$

$(m_1-m_2)=2m_2$

$m_1=3m_2$

The large sphere has to complete 3 times the mass of the sphere1 because it stops after the crash.

b) Let us calculate with the other equation the speed with which the sphere comes out2 (small)

$V_(2f)=(2m_1)/(m_tV_(10)) +(m_2-m_1)/(m_tV_(20))$

$V_(2f)=(2m_1)/(m_tV_h) +(m_2-m_1)/(m_t(-V_h))$

In addition, we know that m₁ = 3 m₂

$m_t=3m_2+m_2$ mt = 3m2 + m2

$m_t=4m_2$

$V_(2f)=(2* 3m_2)/(4m_2V_h-(m_2-3m_2)4m_2V_h)$

$V_(2f)=(3)/(2) V_h+(1)/(2) V_h$

$V_(2f)=2V_h$

$V_(2f)=2\sqrt{2gh$

This is the rate of rising of sphere 2 (small. At the highest point, it zeroes velocity $V_f$ = 0

$V^2=V_(2f)^2-2gy$

$0=(2√(2gh))^2-2gy$

$y=4H$

Thus

a) The large sphere has 3 times the mass of the small sphere

b) The final height at which small ball reach y = 4H

To know more about the Laws of collisions follow

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Answer:

a) the large sphere has 3 times the mass of the small sphere

b) y = 4H

Explanation:

We must start this problem by calculating the speed with which the spheres reach the floor

vf² = vo² - 2g y

As the spheres are released v₀ = 0

vf² = - 2g H

vf = √ (2g H)

The two spheres arrive with the same speed to the floor.

The largest sphere clashes elastically so that with the floor it has a much higher mass, the sphere bounces with the same speed with which it arrived, the exit speed of the spheres

V₁₀ = √2gH

The big sphere goes up and the small one down, the two collide, let's form a system that contains the two spheres, let's use moment conservation

Let's call vh = √2gH

Small sphere m₂ and v₂o = - √2gH = -vh

Large sphere m₁ and v₁o = √ 2gh = vh

Before crash

p₀ = m₁ v₁₀ + m₂ v₂₀

After the crash

pf = m₁ v₁f + m₂ v₂f

po = pf

m₁ v₁₀ + m₂ v₂₀ = m₁ v₁f + m₂ v₂f

The conservation of kinetic energy

Ko = ½ m₁ v₁₀² + ½ m₂ v₂₀²

Kf = ½ m₁ v₁f² + ½ m₂ v₂f²

Ko = KF

½ m₁ v1₁₀² + ½ m₂ v₂₀² = ½ m₁ v₁f² + ½ m₂ v₂f²

Let's write the values

-m₁ vh + m₂ vh = m₁ v₁f + m₂ v₂f

m₁ vh² + m₂ vh² = m₁ v₁f² + m₂ v₂f²

The solution to this system of equations is

mt = m₁ + m₂

v1f = (m₁-m₂) / mt v₁₀ + 2m₂ / mt v₂

v₂f = 2m₁ /mt v₁₀ + (m₂-m₁) / mt v₂

The large sphere is labeled 1, we are asked for the mass so that v1f = 0, let's clear the equation

v₁f = (m₁-m₂) / mt v₁₀ + 2m₂ / mt v₂₀

0 = (m₁-m₂) / mt vh + 2 m₂ / mt (-vh)

(m₁-m₂) / mt vh = 2 m₂ / mt vh

(m₁-m₂) = 2m₂

m₁ = 3 m₂

The large sphere has to complete 3 times the mass of the sphere1 because it stops after the crash.

b) Let us calculate with the other equation the speed with which the sphere comes out2 (small)

v₂f = 2m₁ / mt v₁₀ + (m₂-m₁) / mt v₂₀

v₂f = 2 m₁ / mt vh + (m₂-m₁) mt (-vh)

In addition, we know that m₁ = 3 m₂

mt = 3m2 + m2

mt= 4m2

v₂f = 2 3m₂ / 4m₂ vh - (m₂-3m₂) 4m₂ vh

v₂f = 3/2 vh +1/2 vh

v₂f = 2 vh

v₂f = 2 √ 2gh

This is the rate of rise of sphere 2 (small. At the highest point its zero velocity vf = 0

V² = v₂f² - 2 g Y

0 = (2√2gh)² - 2gy

2gy = 4 (2gH)

y = 4H

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