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Pepsi [2]
3 years ago
11

50 POINTS ANSWER CORRECTLY

Physics
2 answers:
AveGali [126]3 years ago
8 0
I think its c 

Hope this is right
worty [1.4K]3 years ago
3 0
The answer would be B hope this helps :D
You might be interested in
.Which of the following scenarios correctly describes Newton's First Law?
hodyreva [135]

Answer:

A chair that for some reason won't move even after being pushed.

Explanation:

Newton's first law states that, if a body is at rest or moving at a constant speed in a straight line, it will remain at rest or keep moving in a straight line at constant speed unless it is acted upon by a force. This postulate is known as the law of inertia.

Hope this helps! Sorry if I'm wrong.

7 0
3 years ago
Read 2 more answers
An electron moving with a speed of 4.0 * 105 m>s in the positive x direction experiences zero magnetic force. When it moves i
IgorLugansk [536]

Answer:

Explanation:

Given that,

Charge on an electron is

q=1.609×10^-19C

Speed of electron V=4×10^5 m/s in positive x direction

V = 4×10^5 •i m/s

At this point F=0N

When the electron is moving in positive y direction

V =4×10^5 •j m/s

F = 3.2×10^-13 N in positive z direction.

F = 3.2×10 ^-13 •k N

What is the direction and magnitude of the magnetic field B?

The force in a magnetic field is given as

F = q (V ×B)

The only time V×B is zero is when both V and B are in same direction

This shows that B is in the positive x direction or negative x direction

i.e B= Bx •i

Now applying this to the second condition

F= q(V×B)

3.2×10^-13 •k =1.609×10^-19 × (4×10^5•j ×Bx •i)

Note that, j×i =-k

3.2×10^-13 •k / 1.609×10^-19 = (4×10^5Bx •(j×i))

1988812.927 •k =- 4×10^5Bx•k

1988812.927/4×10^5 •k =-Bx •k

4.972 •k =-Bx •k

Then, Bx = -4.972 •k Telsa

Then, the magnitude of B is

B=4.972 T

And its direction is negative z direction.

4 0
3 years ago
Read 2 more answers
If block D weighs 300 lb and block B weighs 275 lb determine the required weight of block C and the angle theta for equilibrium?
adoni [48]
Set up a free body diagram. 

<span>and by reason, Tcd = Tbd </span>

<span>Tbd y = 275 - 300*sinθ </span>
<span>Tcd y = Tc - 300*sin30 </span>

<span>Tbd x = 300*cosθ </span>
<span>Tcdx = 300 * cos30 </span>

<span>Tbd^2 = (275 - 300*sinθ)^2 + (300*cosθ)^2 </span>
<span>Tcd^2 = (300*sin30)^2 + (300 * cos30)^2 </span>

<span>(275 - 300*sinθ)^2 + (300*cosθ)^2 = (300*sin30)^2 + (300 * cos30)^2 </span>
<span>etc.</span>
5 0
3 years ago
Sam, whose mass is 78 kg , stands at the top of a 11-m-high, 110-m-long snow-covered slope. His skis have a coefficient of kinet
Valentin [98]

Answer:

v = 8.09   m/s

Explanation:

For this exercise we use that the work done by the friction force plus the potential energy equals the change in the body's energy.

Let's calculate the energy

       

starting point. Higher

         Em₀ = U = m gh

final point. To go down the slope

         Em_f = K = ½ m v²

The work of the friction force is

         W = fr L cos 180

to find the friction force let's use Newton's second law

Axis y

        N - W_y = 0

        N = W_y

X axis

        Wₓ - fr = ma

let's use trigonometry

        sin  θ = y / L

         sin θ = 11/110 = 0.1

         θ = sin⁻¹  0.1

          θ = 5.74º

         sin 5.74 = Wₓ / W

         cos 5.74 = W_y / W

         Wₓ = W sin 5.74

         W_y = W cos 5.74

the formula for the friction force is

         fr = μ N

         fr = μ W cos θ

Work is friction force is

         W_fr = - μ W L cos θ  

Let's use the relationship of work with energy

        W + ΔU = ΔK

         -μ mg L cos 5.74 + (mgh - 0) = 0  - ½ m v²

        v² = - 2 μ g L cos 5.74 +2 (gh)

        v² = 2gh - 2 μ gL cos 5.74

let's calculate

        v² = 2 9.8 11 - 2 0.07 9.8 110 cos 5.74

        v² = 215.6 -150.16

        v = √65.44

        v = 8.09   m/s

6 0
3 years ago
A Neglecting air resistance, a ball projected straight upward so it remains in the air for 10 seconds needs an initial speed of
ololo11 [35]

Answer:

The initial velocity is 50 m/s.

(C) is correct option.

Explanation:

Given that,

Time = 10 sec

For first half,

We need to calculate the height

Using equation of motion

v^2=u^2+2gh

h =\dfrac{v^2}{2g}....(I)

For second half,

We need to calculate the time

Using equation of motion

h =ut+\dfrac{1}{2}gt_{2}^2

h=0+\dfrac{1}{2}gt_{2}^2

t_{2}=\sqrt{\dfrac{2h}{g}}

Put the value of h from equation (I)

t_{2}=\sqrt{\dfrac{2\times v^2}{g^2}}

t_{2}=\dfrac{v}{g}

According to question,

t_{1}+t_{2}=10

t_{1}=t_{2}

Put the value of t₁ and t₂

\dfrac{v}{g}+\dfrac{v}{g}=10

\dfrac{2v}{g}=10

v=\dfrac{10\times g}{2}

Here, g = 10

The initial velocity is

v=\dfrac{10\times10}{2}

v=50\ m/s

Hence, The initial velocity is 50 m/s.

3 0
3 years ago
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