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Answers

Answer 1

Answer:

Answer:

1kg

Explanation:

this box is the smallest and weighs the least. Hope this helps :]

Related Questions

What advantage is there in using a set of helmholtz coils over just a single small magnet?
Please help really easy
A electromagnetic wave of light has a wavelength of 500 nm. What is the energy (in Joules) of the photon representing the particle interpretation of this light?
Two large parallel metal plates are 1.6 cm apart and have charges of equal magnitude but opposite signs on their facing surfaces. Take the potential of the negative plate to be zero. If the potential halfway between the plates is then +3.8 V, what is the electric field in the region between the plates?
A grandfather clock keeps time using a pendulum consisting of a light rod connected to a small heavy mass. With a rod of length L, the period of oscillation is 2.00 s. What should the length of the rod be for the period of the oscillations to be 1.00 s?

A 0.26 kg rock is thrown vertically upward from the top of a cliff that is 32 m high. When it hits the ground at the base of the cliff, the rock has a speed of 29 m/s. Assuming that air resistance can be ignore and using conservation of mechanical energy, find (a) the initial speed of the rock and (b) the greatest height of the rock as measured from the base of the cliff.

Answers

Answer:

a) The initial speed of the rock is approximately 14.607 meters per second.

b) The greatest height of the rock from the base of the cliff is 42.878 meters.

Explanation:

a) The rock experiments a free-fall motion, that is a vertical uniform accelerated motion due to gravity, in which air friction and effects of Earth's rotation. By Principle of Energy Conservation we have the following model:

$U_(g,1)+K_(1) = U_(g,2)+K_(2)$ (Eq. 1)

Where:

$U_(g,1)$ , $U_(g,2)$ - Initial and final gravitational potential energies, measured in joules.

$K_(1)$ , $K_(2)$ - Initial and final translational kinetic energies, measured in joules.

By definitions of gravitational potential and translational kinetic energies we expand and simplify the equation above:

$m\cdot g\cdot (y_(1)-y_(2))= (1)/(2)\cdot m\cdot (v_(2)^(2)-v_(1)^(2))$

$g\cdot (y_(1)-y_(2)) = (1)/(2)\cdot (v_(2)^(2)-v_(1)^(2))$ (Eq. 2)

Where:

$g$ - Gravitational acceleration, measured in meters per square second.

$y_(1)$ , $y_(2)$ - Initial and final height, measured in meters.

$v_(1)$ , $v_(2)$ - Initial and final speed of the rock, measured in meters per second.

If we know that $g = 9.807\,(m)/(s^(2))$ , $y_(1) = 32\,m$ , $y_(2) = 0\,m$ and $v_(2) = 29\,(m)/(s)$ , then the equation is:

$\left(9.807\,(m)/(s^(2)) \right)\cdot (32\,m-0\,m) = (1)/(2)\cdot \left[\left(29\,(m)/(s) \right)^(2)-v_(1)^(2)\right]$

$313.824 = 420.5-0.5\cdot v_(1)^(2)$

$0.5\cdot v_(1)^(2) = 106.676$

$v_(1) \approx 14.607\,(m)/(s)$

The initial speed of the rock is approximately 14.607 meters per second.

b) We use (Eq. 1) once again and if we know that $g =9.807\,(m)/(s^(2))$ , $y_(1) = 32\,m$ , $v_(1) \approx 14.607\,(m)/(s)$ and $v_(2) = 0\,(m)/(s)$ , then the equation is:

$\left(9.807\,(m)/(s^(2)) \right)\cdot (32\,m-y_(2)) = (1)/(2)\cdot \left[\left(0\,(m)/(s) \right)^(2)-\left(14.607\,(m)/(s) \right)^(2)\right]$

$313.824-9.807\cdot y_(2) = -106.682$

$9.807\cdot y_(2) = 420.506$

$y_(2) = 42.878\,m$

The greatest height of the rock from the base of the cliff is 42.878 meters.

The electric field at the distance of 3.5 meters from an infinite wall of charges is 125 N/C. What is the magnitude of the electric field 1.5 meters from the wall?

Answers

Explanation:

It is given that,

Distance, r = 3.5 m

Electric field due to an infinite wall of charges, E = 125 N/C

We need to find the electric field 1.5 meters from the wall, r' = 1.5 m. Let it is equal to E'. For an infinite wall of charge the electric field is given by :

$E=(\lambda)/(2\pi \epsilon_o r)$

It is clear that the electric field is inversely proportional to the distance. So,

$(E)/(E')=(r')/(r)$

$E'=(Er)/(r')$

$E'=(125* 3.5)/(1.5)$

E' = 291.67 N/C

So, the magnitude of the electric field 1.5 meters from the wall is 291.67 N/C. Hence, this is the required solution.

Two bullets of the same size, mass and horizontal velocity are fired at identical blocks, only one is made of steel and the other is made of rubber. The steel bullet has a perfectly inelastic collision with the block, while the rubber bullet has an elastic collision. Which bullet is more likely to knock over the block, or are both equally likely to do so? Justify your choice based on physics principles.

Answers

Answer and Explanation:

Since we're discussing shots, the significant thing is the way the energy is changed over as there is deceleration of the bullet to a halt when it hits something.

Kinetic Energy is relative to mass times speed squared, so in reality, the 2 cases given have practically indistinguishable Kinetic energy. The measure of energy is authoritative, so the two cases will do generally a similar harm given, obviously we look at situations when all the kinetic energy is spent.

One contrast that will be effectively obvious is that the weapon in the case of heavy bullet will recoil more.
One can consider energy assimilation as force times separation distance, and energy ingestion as a product of force and time.
Henceforth, the heavier yet more slow bullet with a similar energy will venture to every part of a similar separation in the engrossing material, but since of bigger force, will take a more drawn out time doing it.
It will along these lines, additionally, give a more noteworthy "kick" to the object that absorbs.

A 79-turn, 16.035-cm-diameter coil is at rest in a horizontal plane. A uniform magnetic field 43 degrees away from vertical increases from 0.997 T to 6.683 T in 56.691 s. Determine the emf induced in the coil.

Answers

Answer:

The induced emf is 0.0888 V.

Explanation:

Given that,

Number of turns = 79

Diameter = 16.035 cm

Angle = 43

Change in magnetic field $\Delta B=(6.683-0.997)= 5.686\ T$

Time = 56.691 s

We need to calculate the induced emf

Using formula of induced emf

$\epsilon=(NA\Delta B\cos\theta)/(\Delta T)$

Where, N = number of turns

A = area

B = magnetic field

Put the value into the formula

$\epsilon=(79*\pi*(8.0175*10^(-2))^2*5.686*\cos43)/(56.691)$

$\epsilon =0.0888\ V$

Hence, The induced emf is 0.0888 V.

An airplane is traveling 835 km/h in a direction 41.5 ∘ west of north. Find the components of the velocity vector in the northerly and westerly directions. How far north and how far west has the plane traveled after 2.20 h ?

Answers

I assume the graph is looking like in the picture bellow.

North component:
cos(41.5) * 835 = 625.37 km/h

West component of speed:
sin(41.5) * 835 = 553.29 km/h

After 2.2 hours plane will fly:
2.2*625.37 = 1375.81 km north
2.2*553.29 = 1217.23 km west

Final answer:

To find the components of the velocity vector, you can use trigonometry. The north component is calculated using the sine function and the west component is calculated using the cosine function. After 2.20 hours, the distance traveled north and west can be found by multiplying the velocity components by the time.

Explanation:

To find the components of the velocity vector in the northerly and westerly directions, we can use trigonometry. The velocity vector is 835 km/h and is traveling in a direction 41.5° west of north. To find the north component, we can use the sine function: North component = velocity * sin(angle). To find the west component, we can use the cosine function: West component = velocity * cos(angle).

After 2.20 hours, we can find the distance traveled north and west by multiplying the velocity components by the time: Distance north = North component * time and Distance west = West component * time.

Let's calculate the values:

North component = 835 km/h * sin(41.5°)
West component = 835 km/h * cos(41.5°)
Distance north = North component * 2.20 h
Distance west = West component * 2.20 h

Learn more about Velocity components here:

brainly.com/question/14478315

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The graph below shows the velocity of a car as it attempts to set a speedrecord.
Velocity vs. Time
1400
1300
1200
1100
1000
4 45
3
1 (s)
At what point is the car the fastest?
A. t = 1.0 s
B. t = 4.2 s
C. t = 3.0 s
D. t = 4.5 s

Answers

From the graph, it is clear that, the velocity is at a time of 1 s is highest. The velocity at 1 second corresponds to 1250 km/hr. Then it decreases with time.

What is velocity - time graph ?

The velocity - time graph shows the change in velocity with respect to time. The velocity is placed in y -axis and time is given in x - axis. The slope of the curve in velocity - time graph gives the acceleration of the object.

Similarly, the position of the object in meter after a t seconds can be determined from the velocity - time graph. It is the rate of change in velocity of the object.

From the graph, it is clear that, the curve has its peak at 1 second. After that the peak descends down. Hence, the maximum velocity of the car is at a time of 1 second at which the velocity is 1250 km/hr.

Find more on velocity - time graph :

brainly.com/question/28357012

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Guessing (A) because it had the highest velocity number on the graph and it matched 1s

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A ball of mass 0.7 kg flies through the air at low speed, so that air resistance is negligible. (a) What is the net force acting on the ball while it is in motionWhich components of the ball's momentum will be changed by this force? What happens to the x component of the ball's momentum during its flight? What happens to the y component of the ball's momentum during its flight? It decreases. What happens to the z component of the ball's momentum during its flight?

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A smart phone charger delivers charge to the phone, in the form of electrons, at a rate of -0.75. How many electrons are delivered to the phone during 27 min of charging?