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IRISSAK [1]
3 years ago
12

a body of mass 2kg is released from rest on a smooth plane at an angle of 60 degree to the horizontal.calculate the acceleration

of the body down the plane (g=10ms-2​
Physics
1 answer:
Lelu [443]3 years ago
5 0

Answer:

8.66m/s2

Explanation:

from newton second laws F=ma

the force down the plane is the component force along the plane which is mgsin₩ so there fore

a= gsin₩ = 10sin60= 8.66m/s2

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A current is flowing in a wire in direction 3i + 4j where the direction of the magnetic field is 5j + 12k. The force on the wire
nasty-shy [4]

Answer:

F = 0.768 i ^ - 0.576 j ^ + 0.24 k ^

the correct answer is "b"

Explanation:

The magnetic force is

          F = i l x B

The bold are vectors, in this case they give us the direction of the current and the magnetic field, for which we can solve as a determinant

         

F = i \left[\begin{array}{ccc}x&y&z\\3&4&0\\0&5&12\end{array}\right]

resolver

     F = i ^ (4 12 - 0) + j ^ (0- 3 12) + k ^ (3  5 - 0)

     F = i (48 i ^ - 36 j ^ + 15 k⁾

in this case i is the value of the current flowing through the cable

     i = 16 mA = 0.016 A

     F = 0.768 i ^ - 0.576 j ^ + 0.24 k ^

When reviewing the different answers, the correct answer is "b"

6 0
3 years ago
Engineers and science fiction writers have proposed designing space stations in the shape of a rotating wheel or ring, which wou
lyudmila [28]

Answer:

w = 1.976 rpm

Explanation:

For simulate the gravity we will use the centripetal aceleration a_c, so:

a_c = w^2r

where w is the angular aceleration and r the radius.

We know by the question that:

r = 60.5m

a_c = 2.6m/s2

So, Replacing the data, and solving for w, we get:

2.6m/s = w^2(60.5m)

W = 0.207 rad/s

Finally we change the angular velocity from rad/s to rpm as:

W = 0.207 rad/s = 0.207*60/(2\pi)= 1.976 rpm

5 0
3 years ago
The mass of the sun is 2*10^30 kg and its radius is
Marta_Voda [28]

Explanation:

Distance d=1.5×108 km=1.5×1011 m

Mass of the sun, m=2×1030 kg

Mass of the earth, M=6×1024 kg

Force of gravitation, F=G×d2m×M

F=6.7×10−11×(1.5×1011)22×1030×6×1024=3.57×1022 N

4 0
3 years ago
A pulsed proton beam is fired at a target. Each pulse lasts 45.0 ns, and there are protons in each pulse, each proton having a s
skelet666 [1.2K]

Answer:

P = n P₀ 4.9 10¹⁴  Pa

Explanation:

The radiation pressure for full absorption is

        P = S / c

Where S the pointing vector, which is equal to the intensity of the beam that is defined as the energy per unit area per unit time

The energy of the protons can be calculated

       Em = K = ½ m v²

Area

          A = π r²

Intensity is

          I = n ½ m v² / π r²

          I = ½ n m /π v² / r²

We replace

         S = U / t A

          S = ½ n m /π v² / r² Δt

The pressure is

           P = 1/c   (½ n m /π (v / r Δt)²2

           Δt = 45 10⁻⁹ s

           P = n [½ m /πc  (v/r)²] 4.9 10¹⁴

The amount in square brackets is the pressure that a proton creates, which is why it is useful

         P = n P₀ 4.9 10¹⁴  Pa

Where Po is the pressure created by a proton

4 0
4 years ago
At noon, ship A is 140 km west of ship B. Ship A is sailing east at 30 km/h and ship B is sailing north at 25 km/h. How fast (in
Luba_88 [7]

Answer:

The rate of change of distance between the two ships is 18.63 km/h

Explanation:

Given;

distance between the two ships, d = 140 km

speed of ship A = 30 km/h

speed of ship B = 25 km/h

between noon (12 pm) to 4 pm = 4 hours

The displacement of ship A at 4pm = 140 km - (30 km/h x 4h) =

140 km - 120 km = 20 km

(the subtraction is because A is moving away from the initial position and the distance between the two ships is decreasing)

The displacement of ship B at 4pm = 25 km/h x 4h = 100 km

Using Pythagoras theorem, the resultant displacement of the two ships at 4pm is calculated as;

r² = a²   +  b²

r² = 20²  +  100²

r = √10,400

r = 101.98 km

The rate of change of this distance is calculated as;

r² = a²   +  b²

r = 101.98 km, a = 20 km, b = 100 km

2r(\frac{dr}{dt} ) = 2a(\frac{da}{dt} )  + 2b(\frac{db}{dt} )\\\\r(\frac{dr}{dt} ) = a(\frac{da}{dt} )  + b(\frac{db}{dt} )\\\\101.98(\frac{dr}{dt} ) = 20(-30 )  + 100(25 )\\\\101.98(\frac{dr}{dt} ) = -600 + 2,500\\\\101.98(\frac{dr}{dt} ) = 1900\\\\\frac{dr}{dt}  = \frac{1900}{101.98} = 18.63 \ km/h

5 0
3 years ago
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