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Mademuasel [1]
3 years ago
13

A chair of weight 85.0 N lies atop a horizontal floor; the floor is not frictionless. You push on the chair with a force of F =

40.0 N directed at an angle of 35.0deg below the horizontal and the chair slides along the floor.
Using Newton's laws, calculate n, the magnitude of the normal force that the floor exerts on the chair.
Physics
1 answer:
blsea [12.9K]3 years ago
6 0

Answer:

 N = 107.94 N

Explanation:

For this exercise we must use Newton's second law.

Let's set a reference system with the x-axis parallel to the ground and the y-axis vertical

X axis

        Fₓ = ma

ej and

       N -F_y - W = 0

let's use trigonometry to decompose the applied force

     cos -35 = Fₓ / F

     sin -35 = F_y / F

     Fₓ = F cos -35

     F_y = F sin -35

     Fₓ = 40.0 cos -35 = 32.766 N

     F_y = 40.0 sin -35 = -22.94 N

we substitute

     N = Fy + W

     N = 22.94 + 85

     N = 107.94 N

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An electron moving with a velocity = 5.0 × 107 m/s enters a region of space where perpendicular electric and a magnetic fields a
Cloud [144]

Answer:

magnetic field will allow the electron to go through 2 x 10^{4}  T k

Explanation:

Given data in question

velocity = 5.0 × 10^{7}

electric filed = 10^{4}

To find out

what magnetic field will allow the electron to go through, undeflected

solution

we know if electron move without deflection i.e. net force is zero on electron and we can say both electric and magnetic force equal in magnitude and opposite in directions

so we can also say

F(net) = Fe + Fb i.e. = 0

q V B +  q E = 0

q will be cancel out

10^{4}j + 5e + 7i × B =  0

B = 2 x 10^{4}  T k

3 0
3 years ago
At the bottom of its path, the ball strikes a 2.30 kg steel block initially at rest on a frictionless surface. The collision is
Triss [41]

Answer:

(a). The speed of the ball after collision is 2.01 m/s.

(b). The speed of the block after collision 1.11 m/s.

Explanation:

Suppose, A steel ball of mass 0.500 kg is fastened to a cord that is 50.0 cm long and fixed at the far end. The ball is then released when the cord is horizontal.

Given that,

Mass of steel block = 2.30 kg

Mass of ball = 0.500 kg

Length of cord = 50.0 cm

We need to calculate the initial speed of the ball

Using conservation of energy

\dfrac{1}{2}mv^2=mgl

v=\sqrt{2gl}

Put the value into the formula

u=\sqrt{2\times9.8\times50.0\times10^{-2}}

u=3.13\ m/s

The initial speed of the ball u_{1}=3.13\ m/s

The initial speed of the block u_{2}=0

(a). We need to calculate the speed of the ball after collision

Using formula of collision

v_{1}=(\dfrac{m_{1}-m_{2}}{m_{1}+m_{2}})u_{1}+(\dfrac{2m_{2}}{m_{1}+m_{2}})u_{2}

Put the value into the formula

v_{1}=(\dfrac{0.5-2.30}{0.5+2.30})\times3.13

v_{1}=-2.01\ m/s

Negative sign shows the opposite direction of initial direction.

(b). We need to calculate the speed of the block after collision

Using formula of collision

v_{2}=(\dfrac{2m_{1}}{m_{1}+m_{2}})u_{1}+(\dfrac{m_{1}-m_{2}}{m_{1}+m_{2}})u_{2}

Put the value into the formula

v_{2}=(\dfrac{2\times0.5}{0.5+2.30})\times3.13+0

v_{2}=1.11\ m/s

Hence, (a). The speed of the ball after collision is 2.01 m/s.

(b). The speed of the block after collision 1.11 m/s.

8 0
3 years ago
What is the magnite of the net displacement of the mouse?
agasfer [191]

Complete Question

A field mouse trying to escape a hawk runs east for 5.0m, darts southeast for 3.0m, then drops 1.0m down a hole into its burrow. What is the magnitude of the net displacement of the mouse?

Answer:

The  values is  s =  7.49 \  m

Explanation:

From the question we are told that

   The  distance it travels eastward is  x =  5.0 \ m

   The distance it travel towards the southeast  is l  =  3.0\  m

   The distance it travel towards the south is  z =  1 \  m

 

Let x-axis  be east

      y-axis  south

       z-axis into the ground

The angle made between east and south is  \theta  =  45^o

The displacement toward x-axis is

       x =  5 +  3cos(45)

       x =  7.12

 The  displacement toward the y-axis is  

     y  =  3 *  sin (45)

      y =  2.123

Now the overall displacement of the rat is mathematically evaluated as

        s =  \sqrt{7.12^2 +  2.12^2 +  1^2}

      s =  7.49 \  m

     

   

3 0
3 years ago
A 600-g block is dropped onto a relaxed vertical spring that has a spring constant k =190.0 N/m. The block becomes attached to t
charle [14.2K]

Answer:

Work done will be 2.205 j

Explanation:

We have given that the spring is compressed b 37.5 cm

So d = 0.375 m

Mass of the block m = 600 gram = 0.6 kg

Acceleration due to gravity g=9.8m/sec^2

Gravitational force on the block F=mg=0.6\times 9.8=5.88N

Now we know that work done is give by W=Fd=5.88\times 0.375=2.205J

5 0
3 years ago
A bus that was moving at a high speed stopped suddenly. One of
fiasKO [112]

Answer:

this doesn't make sinces

7 0
2 years ago
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