1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
stepladder [879]
2 years ago
15

An open cart is rolling to the left on a horizontal surface. A package slides down a chute and lands in the cart. Which quantity

or quantities have the same value just before and just after the package lands in the cart
Physics
1 answer:
Marat540 [252]2 years ago
8 0

Answer: the horizontal component of total momentum

Explanation:

Since the open cart is rolling to the left on the horizontal surface, the quantity that has the same value just before and just after the package lands in the cart is the horizontal component of total momentum.

Momentum, is the product of the mass of a particle and the velocity of the particle. The change of momentum depends on the force which acts on it. The addition of the the individual momenta is the total momentum.

You might be interested in
suppose a car manufacturer tested its cars for front end collsion by hauling them up on a crane and dropping them from a certain
IRINA_888 [86]

Initial height: 66.5 m

Explanation:

The problem can be solved by using the principle of conservation of energy.

If we neglect air resistance, the total mechanical energy of the car is conserved during the fall, therefore we can write:

K_i + U_i = K_f + U_f

where :

K_i = 0 is the kinetic energy of the car at the top (it starts from rest)

U_i = mgh is the gravitational potential energy of the car at the top, with:

m = the mass of the car

g = the acceleration of gravity

h = the heigth of the car

K_f = \frac{1}{2}mv^2 is the kinetic energy of the car just before hitting the ground, with

v = 130 km/h final speed of the car

U_f = 0 is the gravitational potential energy of the car at the bottom

Re-arranging the equation,  we find

mgh=\frac{1}{2}mv^2

and we have:

g=9.8 m/s^2\\v = 130 km/h = 36.1 m/s

Solving for h, we find the initial height of the car:

h=\frac{v^2}{2g}=\frac{36.1^2}{2(9.8)}=66.5 m

Learn more about kinetic energy and potential energy:

brainly.com/question/6536722

brainly.com/question/1198647  

brainly.com/question/10770261  

#LearnwithBrainly

5 0
3 years ago
Tom rides his motorcycle at a speed of 15 meters/second for an hour.
Dahasolnce [82]

Answer:

1 hour to ride his motorcycle

4 0
2 years ago
7. A beam of light converges at a point P. Now a lens is placed in the path of the convergent
Rudiy27

Answer:

Lens at a distance = 7.5 cm

Lens at a distance = 6.86 cm  (Approx)

Explanation:

Given:

Object distance u = 12 cm

a) Focal length = 20 cm

b) Focal length = 16 cm

Computation:

a. 1/v = 1/u + 1/f

1/v = 1/20 + 1/12

v = 7.5 cm

Lens at a distance = 7.5 cm

b.  1/v = 1/u + 1/f

1/v = 1/16 + 1/12

v = 6.86 cm  (Approx)

Lens at a distance = 6.86 cm  (Approx)

5 0
3 years ago
What can happen to solar radiation when it enters atmosphere​
grandymaker [24]

Answer:

The solar radiation is first intercepted by Earth's atmosphere, just a small part of the radiation is absorbed by gases such as water vapor. Some of the radiation is reflected back to space by the clouds and Earth's surface.

3 0
3 years ago
A 41-turn square coil of area 0.074 m2 and a 123-turn circular coil are both placed perpendicular to the same changing magnetic
vesna_86 [32]

Answer:

<h3>The area of second coil is ≅ 0.025 m^{2}</h3>

Explanation:

Given :

No. of turns in the first coil N_{1} = 41

No. of turns in the second coil N_{2}  = 123

Area of first coil A_{1} = 0.074 m^{2}

According to the law of electromagnetic induction,

Induced emf = -N \frac{d \phi}{dt}

Where \phi = magnetic flux.

Since given in question emf of both coil is same so we compare above equation.

    -\frac{N_{1} d\phi _{1}   }{dt_{1} }  = -\frac{N_{2} d\phi _{2}   }{dt_{2} }

   \frac{N_{1} A_{1}   dB_{1}  }{dt_{1} }  = \frac{N_{2} A_{2} dB_{2}     }{dt_{2} }

        A_{2} = \frac{N_{1} A_{1}  }{N _{2}  }

        A_{2} = \frac{41 \times 0.074 }{123  }

        A_{2} = 0.0246 = 0.025 m^{2}

Therefore, the area of second coil is ≅ 0.025 m^{2}

4 0
3 years ago
Other questions:
  • "5 N, up" is an example of a ___.<br> OA) force<br> OB) mass<br> OC) weight<br> OD) magnitude
    14·1 answer
  • An arrow is used to show a force acting on an object. what information does the arrow give about the force
    6·1 answer
  • Which structure is part of the inner ear?
    12·2 answers
  • In a compound chemical energy is contained in the what ?
    9·2 answers
  • Forces always act in pairs? True or false
    8·1 answer
  • What is the most problem of any defective tools?​
    6·1 answer
  • Determine the force of gravitational attraction between the Earth and the moon. Their masses are 5.98 x 1024 kg and 7.26 x 1022
    7·1 answer
  • 9.
    13·1 answer
  • Darius' boat sails into the harbor with a speed of 80m/s. After 20 seconds, Darius' boat has come to a stop at the dock. What is
    14·1 answer
  • Please help me with part 2. Will give brainliest
    6·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!