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
ziro4ka [17]
3 years ago
7

Hii please help i’ll give brainliest!!

Physics
1 answer:
faltersainse [42]3 years ago
8 0
The mass of the box
HOPE THIS HELPS❤️
You might be interested in
A train at a constant 79.0 km/h moves east for 27.0 min, then in a direction 50.0° east of due north for 29.0 min, and then west
ivolga24 [154]

Answer:

Magnitude of avg velocity, |v_{avg}| = 18.9 km/h

\theta' = 56.85^{\circ}

Given:

Constant speed of train, v = 79 km/h

Time taken in East direction, t = 27 min = \frac{27}{60} h

Angle, \theta = 50^{\circ}

Time taken in 50^{\circ}east of due North direction, t' = 29 min =  \frac{29}{60} h

Time taken in west direction, t'' = 37 min =  \frac{27}{60} h

Solution:

Now, the displacement, 's' in east direction is given by:

\vec{s} = vt = 79\times \frac{27}{60} = 35.5\hat{i} km

Displacement in  50^{\circ} east of due North for 29.0 min is given by:

\vec{s'} = vt'sin50^{\circ}\hat{i} + vt'cos50^{\circ}\hat{j}

\vec{s'} = 79(\frac{29}{60})sin50^{\circ}\hat{i} + 79(\frac{29}{60})cos50^{\circ}\hat{j}

\vec{s'} = 29.25\hat{i} + 24.54\hat{j} km

Now, displacement in the west direction for 37 min:

\vec{s''} = - vt''hat{i} = - 79\frac{37}{60} = - 48.72\hat{i} km

Now, the overall displacement,

\vec{s_{net}} = \vec{s} + \vec{s'} + \vec{s''}

\vec{s_{net}} = 35.5\hat{i} + 29.25\hat{i} + 24.54\hat{j} - 48.72\hat{i}

\vec{s_{net}} =  16.03\hat{i} + 24.54\hat{j} km

(a) Now, average velocity, v_{avg} is given:

v_{avg} = \frac{total displacement, \vec{s_{net}}}{total time, t}

v_{avg} = \frac{16.03\hat{i} + 24.54\hat{j}}{\frac{27 + 29 + 37}{60}}

v_{avg} = 10.34\hat{i} + 15.83\hat{j}) km/h

Magnitude of avg velocity is given by:

|v_{avg}| = \sqrt{(10.34)^{2} + (15.83)^{2}} = 18.9 km/h

(b) angle can be calculated as:

tan\theta' = \frac{15.83}{10.34}

\theta' = tan^{- 1}\frac{15.83}{10.34} = 56.85^{\circ}

6 0
3 years ago
Determine the magnitude of the force between two 42 m-long parallel wires separated by 0.03 m, both carrying 6.3 A in the same d
xz_007 [3.2K]

Answer:

The magnitude of the force between the two parallel wires is 0.0111 N.

Explanation:

Given;

length of the two parallel wires, L = 42 m

distance between the two wires, r = 0.03 m

current in both wires, I₁, I₂ = 6.3 A

Therefore, the magnitude of the repulsive force between the two parallel wires is given by;

F = \frac{\mu_0 I_1I_2l}{2\pi r}\\\\where;\\\mu_0 \ is \ permeability \ of \ free \ space = 4\pi *10^{-7} \ T.m/A \\\\F = \frac{(4\pi *10^{-7})(6.3)^2(42)}{2\pi (0.03)}\\\\F =   0.0111 \ N

Therefore, the magnitude of the force between the two parallel wires is 0.0111 N.

6 0
3 years ago
A straight wire segment 5 m long makes an angle of 30° with a uniform magnetic field of 0.37 T. Find the magnitude of the force
SIZIF [17.4K]

Answer:

The magnitude of the force on the wire is 2.68 N.

Explanation:

Given that,

Length of the wire, L = 5 m

Magnetic field, B = 0.37 T

Angle between wire and the magnetic field, \theta=30^{\circ}

Current in the wire, I = 2.9 A

We need to find the magnitude of the force on the wire. The magnetic force in the wire is given by :

F=BIL\ \sin\theta\\\\F=0.37\ T\times 2.9\ A\times 5\ m\times \ \sin(30)\\\\F=2.68\ N

So, the magnitude of the force on the wire is 2.68 N. Hence, this is the required solution.

7 0
3 years ago
What is the energy level of hydrogen?
barxatty [35]

Answer:

The energies corresponding to each of the allowed orbitals are called energy levels.

Explanation:

A scientist known as Niels Bohr put forward that electrons in an atom covers some permitted orbitals with a specific energy. In other words, the energy of an electron in an atom is not continuous, but 'quantized.' The energies corresponding to each of the allowed orbitals are called energy levels.

E = -\frac{E_0}{n^2} \\where \\E_0 = 13.6 eV (1 eV = 1.602\times 10^{-19}Joules)\\and\ n = 1,2,3...

8 0
3 years ago
On a playground slide, a child has potential energy that decreases by 1000 J while her kinetic energy increases by 900 J. what o
mel-nik [20]

Answer:

Friction     100 j

Explanation:

Friction causes the inequality between PE and KE

100 J

8 0
2 years ago
Other questions:
  • A 52-kg bike is moving along a smooth road at a constant velocity of 8.4 m/s. What is the net force acting on the bike
    15·1 answer
  • Which of the following stages follows a protostar?
    13·1 answer
  • PLEASE HELP!!! URGENT MATTERS!!!!
    13·1 answer
  • Coulomb is a very large unit for practical use. Justify your answer if 10^10 electrons are transferred from a body/second
    10·1 answer
  • The power radiated by the sun is 3.90 1026 W. The earth orbits the sun in a nearly circular orbit of radius 1.50 1011 m. The ear
    7·1 answer
  • A salad including lettuce , tomatoes , and onions would be classified as a
    10·2 answers
  • Energy cannot be created nor destroyed, only changed from one form to another. How does listening to music on a radio obey the l
    15·1 answer
  • Mr Yadav buys a radio for Rs 640 and sells it for Rs 672 Find it's profit<br><br>​
    5·1 answer
  • Two spherical objects with a mass of 3.17 kg each are placed at a distance of 2.96 m apart. How many electrons need to leave eac
    14·1 answer
  • A 5.0 kg cannonball is fired horizontally at 68 m/s from a 15-m-high cliff. A strong tailwind exerts a constant 12 N horizontal
    14·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!