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3 years ago

A 5.0-kg bag of groceries is tossed onto a table at 2.0 m/s and slides to a stop in 1.6 s .

Physics

1 answer:

son4ous [18]3 years ago

5 0

Answer:

-6.3 N

Explanation:

The equation we need to use is:

$F \Delta t = \Delta (mv)$

where

F is the force of friction, that slows down the bag

$\Delta t$ is the time of the motion

m is the mass of the bag

v is the speed of the bag

Since the mass of the bag does not change, we can rewrite the equation as

$F \Delta t= m \Delta v$

where we have:

$\Delta t=1.6 s\\m=5.0 kg\\\Delta v=-2.0 m/s$

Substituting and re-arranging the equation, we can find the force of friction:

$F=\frac{m\Delta v}{\Delta t}=\frac{(5.0 kg)(-2.0 m/s)}{1.6 s}=-6.3 N$

where the negative sign means that the force is in the opposite direction to the motion of the bag.

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An alpha particle (charge +2e) travels in a circular path of radius .5m in a magnetic field of 1.0 T. Find the (a) period, (b) s

navik [9.2K]

Given Information:

Radius = r = 0.5 m

Magnetic field = 1.0 T

Required Information:

Period = T = ?

Speed = v = ?

Kinetic energy = KE = ?

Answer:

Period = 0.13x10⁻⁶ seconds

speed = 24.16x10⁶ m/s

Kinetic energy = 12.11 MeV

Explanation:

(a) period

The time period of alpha particle is related to its orbital speed as

T = 2πr/v eq. 1

According to newton's law

F = ma

Force due to magnetic field is given by

F = qvB

qvB = ma

qvB = m(v²/r)

qB = mv/r

v = qBr/m eq. 2

substitute the eq. 2 in eq. 1

T = 2πr/qBr/m

r cancels out

T = 2π/qB/m

T = 2πm/qB

T = 2π*6.65x10⁻²⁷/2*1.602x10⁻¹⁹*1

T = 0.13x10⁻⁶ seconds

(b) speed

From equation 1

T = 2πr/v

v = 2πr/T

v = 2π*0.5/0.13x10⁻⁶

v = 24.16x10⁶ m/s

Kinetic energy is given by

KE = 0.5mv²

KE = 0.5*6.65x10⁻²⁷*(24.16x10⁶)²

KE = 1.94x10⁻¹² J

since 1 electron volt has 1.602x10⁻¹⁹ J

KE = 1.94x10⁻¹²/1.602x10⁻¹⁹

KE = 12.11 MeV

5 0

3 years ago

I am sitting on a train car traveling horizontally at a constant speed of 50 m/s. I throw a ball straight up into the air.

drek231 [11]

Answer:

Explanation:

I am sitting on a train car traveling horizontally at a constant speed of 50 m/s. I throw a ball straight up into the air. Before , the ball gets separated from my hand , both me the ball will be moving with velocity of 50 m /s in horizontal direction .

As soon as ball is separated from the hand , it acquires addition velocity in upward direction and acceleration in downward direction . This will give relative velocity to the ball with respect to me . So I will see the ball going in upward direction under gravitational acceleration . It appears as if I am sitting at rest and ball is going in upward direction under deceleration . My motion at 50 m/s will have no effect on the motion of ball in upward direction , according to first law of Newton . It is so because ball too will be moving in forward direction with the same speed which will not be visible to me because I too am moving with the same speed.

If I am sitting at rest at home and I threw a ball straight up into the air , I will have the same experience of seeing ball going in similar way as described above.

8 0

3 years ago

To test the quality of a tennis ball, you drop it onto the floor from a height of 4.00 m. It rebounds to a height of 2.00 m. If

arlik [135]

Answer:

Part a)

$a = 1260.3 m/s^2$

Part b)

Direction = upwards

Explanation:

When ball is dropped from height h = 4.0 m

then the speed of the ball just before it will strike the ground is given as

$v_f^2 - v_i^2 = 2 a d$

$v_1^2 - 0^2 = 2(9.81)(4.0)$

$v_1 = 8.86 m/s$

Now ball will rebound to height h = 2.00 m

so the velocity of ball just after it will rebound is given as

$v_f^2 - v_i^2 = 2 a d$

$0 - v_2^2 = 2(-9.81)(2.00)$

$v_2 = 6.26 m/s$

Part a)

Average acceleration is given as

$a = \frac{v_f - v_i}{\Delta t}$

$a = \frac{6.26 - (-8.86)}{12.0 \times 10^{-3}}$

$a = 1260.35 m/s^2$

Part B)

As we know that ball rebounds upwards after collision while before collision it is moving downwards

So the direction of the acceleration is vertically upwards

7 0

3 years ago

A 12,000kg. railroad car is traveling at +2m/s when it

ivann1987 [24]

Answer:

The final velocity of the two railroad cars is 1.09 m/s

Explanation:

Since we are given that the two cars lock together it shows that the collision is inelastic in nature. The final velocity due to inelastic collision is given by

$\mathrm{V}=\frac{V 1 M 1+V 2 M 2}{M 1+M 2}$

where

V= Final velocity

M1= mass of the first object in kgs = 12000

M2= mas of the second object in kgs = 10000

V1= initial velocity of the first object in m/s = 2m/s

V2= initial velocity of the second object in m/s = 0 (given at rest)

Substituting the given values in the formula we get

V = 2×12000 + 0x100012000 + 10000= 2400022000= 1.09 m/s

$\mathrm{V}=\frac{2 \times 1200+0 \times 1000}{12000+10000}=\frac{24000}{22000}=1.09 \mathrm{m} / \mathrm{s}$

Which is the final velocity of the two railroad cars

8 0

3 years ago

All of the following involve waves of electromagnetic energy except _______.

r-ruslan [8.4K]

All of the following involve waves of electromagnetic energy except the rumble of thunder during a storm.

Electromagnetic waves are used to transmit long/short/FM wavelength radio waves, and TV/telephone/wireless signals or energies. They are also responsible for transmiting energy in the form of microwaves, infrared radiation (IR), visible light (VIS), ultraviolet light (UV), X-rays, and gamma rays.

The correct answer between all the choices given is the second choice or letter B. I am hoping that this answer has satisfied your query and it will be able to help you in your endeavor, and if you would like, feel free to ask another question.

6 0

3 years ago

Read 2 more answers