PHYSICS COLLEGE

A revolutionary war cannon, with a mass of 2260 kg, fires a 21 kg ball horizontally. The cannonball has a speed of 105 m/s after it has left the barrel. The cannon carriage is on a flat platform and is free to roll horizontally. What is the speed of the cannon immediately after it was fired?

Answers

Answer 1

Answer:

Answer:

0.97566 m/s

Explanation:

$m_1$ = Mass of cannon = 2260 kg

$v_1$ = Velocity of cannon

$m_2$ = Mass of ball = 21 kg

$v_2$ = Velocity of ball = 105 m/s

As the momentum of the system is conserved we have

$m_1v_1=m_2v_2\n\Rightarrow v_1=(21* 105)/(2260)\n\Rightarrow v_1=0.97566\ m/s$

The velocity of the cannon is 0.97566 m/s

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A 100-N uniform ladder, 8.0 m long, rests against a smooth vertical wall. The coefficient of static friction between ladder and floor is 0.40. What minimum angle can the ladder make with the floor before it slips?

Answers

The minimum angle that the ladder make with the floor before it slips is 51.34 Degree.

Given data:

The weight of ladder is, W = 100 N.

The length of ladder is, L = 8.0 m.

The coefficient of static friction between ladder and floor is, $\mu =0.40$ .

Apply the Newton' law in vertical direction to obtain the value of Normal Force (P) as,

$N = mg$

And force along the horizontal direction is,

$F= \mu * N\n\nF = \mu * mg$

Now, taking the torque along the either end of ladder as,

$-mgcos \theta * (L)/(2)+Fsin \theta * L =0\n\nmgcos \theta * (L)/(2) = Fsin \theta * L$

Solving as,

$mgcos \theta * (L)/(2) = (\mu mg) * sin \theta * L\n\ncos \theta * (1)/(2) = (\mu) * sin \theta\n\ntan \theta = (1)/(2 * 0.40 )\n\n\theta = tan^(-1)(1/0.80)\n\n\theta = 51.34^(\circ)$

Thus, we can conclude that the minimum angle that the ladder make with the floor before it slips is 51.34 Degree.

Learn more about the frictional force here:

brainly.com/question/4230804

Answer:

The minimum angle is 51.34°

Explanation:

Given that,

Weight of ladder = 100 N

Length = 8.0 m

Coefficient of static friction = 0.40

We need to calculate the normal force

Using Newtons law in vertical direction

$F_(y)=n-mg$

$N-mg=0$

$N=mg$

We need to calculate the normal force

Using Newtons law in horizontal direction

$F_(s)=f_(s)-P$

$f_(s)-P=0$

$f_(s)=P$

$P=\mu mg$

We need to calculate the minimum angle

Using torque about the point A then

$-mg\cos\theta* AB+P\sin\theta* AC=0$

Put the value into the formula

$mg\cos\theta*((L)/(2))=\mu mg\sin\theta* L$

$\cos\theta*(1)/(2)=\mu\sin\theta$

$(1)/(2)=\mu*\tan\theta$

$\theta=\tan^(-1)((1)/(2*0.40))$

$\theta=51.34^(\circ)$

Hence, The minimum angle is 51.34°

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

Halogen lightbulbs allow their filaments to operate at a higher temperature than the filaments in standard incandescent bulbs. For comparison, the filament in a standard lightbulb operates at about 2900K, whereas the filament in a halogen bulb may operate at 3400K. Which bulb has the higher peak frequency? Calculate the ratio of the peak frequencies. The human eye is most sensitive to a frequency around 5.5x10^14 Hz. Which bulb produces a peak frequency close to this value?

Answers

Answer:

Halogen

0.85294

Explanation:

c = Speed of light = $3* 10^8\ m/s$

b = Wien's displacement constant = $2.897* 10^(-3)\ mK$

T = Temperature

From Wien's law we have

$\lambda_m=(b)/(T)\n\Rightarrow \lambda_m=(2.897* 10^(-3))/(2900)\n\Rightarrow \lambda_m=9.98966* 10^(-7)\ m$

Frequency is given by

$\nu=(c)/(\lambda_m)\n\Rightarrow \nu=(3* 10^8)/(9.98966* 10^(-7))\n\Rightarrow \nu=3.00311* 10^(14)\ Hz$

For Halogen

$\lambda_m=(b)/(T)\n\Rightarrow \lambda_m=(2.897* 10^(-3))/(3400)\n\Rightarrow \lambda_m=8.52059* 10^(-7)\ m$

Frequency is given by

$\nu=(c)/(\lambda_m)\n\Rightarrow \nu=(3* 10^8)/(8.52059* 10^(-7))\n\Rightarrow \nu=3.52088* 10^(14)\ Hz$

The maximum frequency is produced by Halogen bulbs which is closest to the value of $5.5* 10^(14)\ Hz$

Ratio

$(3.00311* 10^(14))/(3.52088* 10^(14))=0.85294$

The ratio of Incandescent to halogen peak frequency is 0.85294

An electromagnetic wave is traveling the +y direction. The maximum magnitude of the electric field associated with the wave is Em, and the maximum magnitude of the magnetic field associated with the wave is Bm. At one instant, the electric field has a magnitude of 0.25 Em and points in the +x direction. At this same instant, the wave's magnetic field has a magnitude which is ... and it is in the ... direction.

Answers

Answer:

0.83x10^-9 T

Direction is towards +z axis.

Explanation:

E = cB

E = magnitude of electrical 0.25 Em

c = speed of light in a vacuum 3x10^8 m/s

Therefore,

B = E/c = 0.25 ÷ 3x10^8

B = 0.83x10^-9 T

Magnetic fueld of a EM wave acts perpendicularly to its electric field, therefore it's direction is towards the +Z axis

At the instant the traffic light turns green, a car that has been waiting at an intersection starts ahead with a constant acceleration of 3.60 m/s^2 . At the same instant a truck, traveling with a constant speed of 23.5 m/s , overtakes and passes the car. a. How far beyond its starting point does the car overtake the truck?b. How fast is the car traveling when it overtakes the truck?

Answers

Answer:

306.8264448 m

47.0016 m/s

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration

Distance traveled by car

$s_c=ut+(1)/(2)at^2\n\Rightarrow s_c=(1)/(2)at^2$

Distance traveled by truck

$s_t=ut$

In order to overtake both distances should be equal

$(1)/(2)at^2=ut\n\Rightarrow (1)/(2)at=u\n\Rightarrow t=(2u)/(a)\n\Rightarrow t=(2* 23.5)/(3.6)\n\Rightarrow t=13.056\ s$

$s_c=(1)/(2)at^2\n\Rightarrow s_c=(1)/(2)3.6* 13.056^2\n\Rightarrow s_c=306.8264448\ m$

The distance the car has to travel is 306.8264448 m

$v^2-u^2=2as\n\Rightarrow v=√(2as+u^2)\n\Rightarrow v=√(2* 3.6* 306.8264448+0^2)\n\Rightarrow v=47.0016\ m/s$

The speed of the car when it overtakes the truck is 47.0016 m/s

A space probe is launched from Earth headed for deep space. At a distance of 10,000 miles from Earth's center, the gravitational force on it is 600lb, what is the size of the force when it is at each of the following distances from the Earths surface?a. 20,000 miles
b. 30,000 miles
c. 100,000 miles

Answers

Grav Force = GMm/r squared. M is earth's mass. m is the probe. r is distance to centre of earth.Given that 600 = GMm/(10,000^2).GMm = 600x(10,000^2) for 10, 000 milesGMm = ? x (20,000^2) for 20, 000 milestherefore600x(10,000^2)=? x (20,000^2)therefore [600x(10,000^2)]/(20,000^2)=? [600x(1x1)]/(2x2)=? [600/4]=? ?=150lb.Same procedure for b and c.Please ask if you want further help.

gravitational force varies based on 1/r^2

when you're double the distance =10,000 to 20,000, the force is 4 times smaller so on and so forth.

As force is proportional to 1 / {distance squared}, the force will be 1 / 2^2 (i.e. 1/4) of the force at the reference distance (i.e. 1/4 * 600 = 150 lb)

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