PHYSICS COLLEGE

A swimmer heads directly across a river, swimming at 1.00 m/s relative to still water. He arrives at a point 41.0 m downstream from the point directly across the river, which is 73.0 m wide. What is the speed of the river current?

Answers

Answer 1
Answer:

Answer:

velocity of the river is equal to 0.56 m/s

Explanation:

given,

velocity of swimmer w.r.t still water = 1 m/s

width of river = 73 m

he arrives to the point = 41 m

times = (distance)/(speed)      

times = (73)/(1)          

 t = 73 s                        

velocity = (distance)/(time)                  

                    = (41)/(73)                      

                    = 0.56 m/s                        

velocity of the river is equal to 0.56 m/s

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A rubber ball is shot straight up from the ground with speed v0. Simultaneously, a second rubber ball at height h directly above the first ball is dropped from rest. a) At what height above the ground do the balls collide? Your answer will be a symbolic expression in terms of v0, h, and g .b) What is the maximum value of h for which a collision occurs before the first ball falls back to the ground?Express your answer in terms of the variable v0 and appropriate constants.c) For what value of h does the collision occur at the instant when the first ball is at its highest point?Express your answer in terms of the variable v0 and appropriate constants.
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How many joules of work are done on an object when a force of 10 N pushes it 5 m?A) 2 JB) 5 JC) 50 JD) 1 JE) 10 J
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A propeller is modeled as five identical uniform rods extending radially from its axis. The length and mass of each rod are 0.777 m and 2.67 kg, respectively. When the propellor rotates at 573 rpm (revolutions per minute), what is its rotational kinetic energy?

Answers

The formula for the rotational kinetic energy is

KE_(rot) = (1)/(2)(number \ of\ propellers)( I)( omega)^(2)

where I is the moment of inertia. This is just mass times the square of the perpendicular distance to the axis of rotation. In other words, the radius of the propeller or this is equivalent to the length of the rod. ω is the angular velocity. We determine I and ω first.

I=m L^(2)=(2.67 \ kg) (0.777 \ m)^(2) =2.07459 \ kgm^(2)

ω = 573 rev/min * (2π rad/rev) * (1 min/60 s) = 60 rad/s

Then,

KE_(rot) =( (1)/(2) )(5)(2.07459 \ kgm^(2)) (60\ rad/s)^(2)

KE_(rot) =18,671.31 \ J

Answer:

4833J

Explanation:

Length=0.777

mass=2.67

# rods= 5

ω=573 rpm--> 573*2\pi *(1)/(60) =60rad/s

I=(1)/(3) mL^2=(1)/(3) (2.67kg)(0.777m)^2=0.537kgm^2

K=1/2(number of rods)(I)(ω)=(1)/(2) *(5)(0.537)(60)^2=4833J

I know it's very late, but hope this helps anyone else trying to find the answer.

If you want to make a strong battery, should you pair two metals with high electron affinities, low electron affinities, or a mix? Explain your answer.

Answers

Answer:

Mix

Explanation:

A battery has two electrodes at both of its end terminals namely the anode which is the negatively charged electrode and the anode which is the positively charged electrode.

Now, Electrons usually travel through the battery circuit from the anode to the cathode, and this motion of travel is the propelling force that makes electricity flow through the circuit.

Now, the anode will need to have a low electron affinity because it needs to easily release electrons during discharge while the cathode needs to have a high electron affinity because it normally readily accept electrons during discharge.

Thus, for a battery to be strong, it is a combination of high electron affinity and low electron affinity.

Answer: mix

Explanation:

you cant pair 2 of the same electron affinities

State the following forms of electromagnetic radiation in increasing order of wavelength.Radiowaves, gamma rays, x-rays, infrared radiation, visible light​

Answers

Answer:

Gamma rays, x-rays, visible light, infrared radiation and radiowaves

Explanation:

Gamma rays, x-rays, ultraviolet, visible light, infrared radiation, microwave and radiowaves

When running a 100 meter race Wyatt reaches his maximum speed when he is 40 meters from the starting line, and 7 seconds have elapsed since the start of the race. Wyatt continues at this max speed for the rest of the race and is 85 meters from the starting line 12 seconds after the start of the race. What is Wyatt's max speed

Answers

Answer:

9 m/s

Explanation:

Wyatt maintains the maximum speed for the rest of the race. This motion begins when his displacement is 40 m and the time is 7 s. At time 12 s, his displacement is 85 m. Because this motion is constant-velocity, the maximum speed is given by

v_\text{max} = (85-40)/(12-7) = (45)/(5) = 9 \text{ m/s}

Calculate the work done by a 50.0 N force on an object as it moves 9.00 m, if the force is oriented at an angle of 135° from the direction of travel. Explain what your answer means in terms of the object’s energy.

Answers

Answer:

Work done, W = -318.19 Joules

Explanation:

It is given that,

Force acting on the object, F = 50 N

Distance covered by the force, d = 9 m

Angle between the force and the distance traveled, \theta=135^(\circ)

The work done by an object is equal to the product of force and distance traveled. It is equal to the dot product of force and the distance. Mathematically, it is given by :

W=Fd\ cos\theta

W=50* 9* \ cos(135)

W = -318.19 Joules

So, the work done by the force is 318.19 Joules. The work is done in opposite to the direction of motion. Hence, this is the required solution.

The brakes of a car moving at 14m/s are applied, and the car comes to a stop in 4s. (a) What was the cars acceleration? (b) How long would the car take to come to a stop starting from 20m/s with the same acceleration? (c) How long would the car take to slow down from 20m/s to 10m/s with the same acceleration?

Answers

(1) The acceleration of the car will be a=-3.5(m)/(s^2)

(2) The time taken t=5.7s

(3)  The time is taken by the car  to slow down from 20m/s to 10m/s t=2.9s

What will be the acceleration and time of the car?

(1) The acceleration of the car will be calculated as

a=(v-u)/(t)

Here

u= 14 (m)/(s)

a=(0-14)/(4) =-3.5(m)/(s^2)

(2) The time is taken for the same acceleration to 20(m)/(s)

a=(v-u)/(t)

t=(v-u)/(a)

u=20(m)/(s)

t=(0-20)/(-3.5) =5.7s

(3) The time is taken to slow down from 20m/s to 10m/s with the same acceleration

From same formula

t=(v-u)/(a)

v=10(m)/(s)

u=20(m)/(s)

t=(10-20)/(-3.5) =2.9s

Thus

(1) The acceleration of the car will be a=-3.5(m)/(s^2)

(2) The time taken t=5.7s

(3)  The time is taken by the car  to slow down from 20m/s to 10m/s t=2.9s

To know more about the Equation of the motion follow

brainly.com/question/25951773

(a) -3.5 m/s^2

The car's acceleration is given by

a=(v-u)/(t)

where

v = 0 is the final velocity

u = 14 m/s is the initial velocity

t = 4 s is the time elapsed

Substituting,

a=(0-14)/(4)=-3.5 m/s^2

where the negative sign means the car is slowing down.

(b) 5.7 s

We can use again the same equation

a=(v-u)/(t)

where in this case we have

a=-3.5 m/s^2 is again the acceleration of the car

v = 0 is the final velocity

u = 20 m/s is the initial velocity

Re-arranging the equation and solving for t, we find the time the car takes to come to a stop:

t=(v-u)/(a)=(0-20)/(-3.5)=5.7 s

(c) 2.9 s

As before, we can use the equation

a=(v-u)/(t)

Here we have

a=-3.5 m/s^2 is again the acceleration of the car

v = 10 is the final velocity

u = 20 m/s is the initial velocity

Re-arranging the equation and solving for t, we find

t=(v-u)/(a)=(10-20)/(-3.5)=2.9 s
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