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Shtirlitz [24]
3 years ago
10

Need help with number 1

Physics
1 answer:
Margarita [4]3 years ago
3 0
200 + 175 + 125 + 100 = 600
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A charged particle isinjected into a uniform magnetic field such that its velocity vector is perpendicular to themagnetic field
Bumek [7]

Answer:

D) follows a circular path

Explanation:

This is because the magnetic force F on the charge, q due to the magnetic field B with velocity vector v perpendicular to it, equals the centripetal force acting on the charge.

So, F = Bqv = mv²/r.

So it follows a circular path.

4 0
4 years ago
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A 2-kg bowling ball rolls at a speed of 10 m/s on the ground.
liberstina [14]
It's kinetic energy as the ball the ball isn't raised above the ground it does not have any gravitational potential energy. 
To find the kinetic energy of the ball you will have to use the formula:
KE=0.5 x m x v squared
m being mass and v being velocity 
so the calculation is:
0.5 x 2 x 10 x 10= 100J
8 0
4 years ago
Please Help!!! It's for a quiz!!
Sphinxa [80]

Answer:

38.8 m/s

Explanation:

Force F(x) = 6 - 2x + 6x²

work

W=\displaystyle\int_{0}^{13.9}F(x)dx=\displaystyle\int_{0}^{13.9}(6-2x+6x^2)dx

=6x-x^2+2x^3|_{0}^{13.9}\\=5261 J

W = mv²/2=7v²/2 = 3.5v² = 5261

v = 38.8 m/s

3 0
2 years ago
A 200 kg weather rocket is loaded with 100 kg of fuel and fired straight up. It accelerates upward at 35 m/s2 for 35 s , then ru
r-ruslan [8.4K]

Answer:

T = 295.57 s

Explanation:

given,

mass of the rocket = 200 Kg

mass of the fuel = 100 Kg

acceleration = 35 m/s²

time, t = 35 s

time taken by the rocket to hit the ground, = ?

Final speed of the rocket when fuel is empty

using equation of motion

v = u + a t

v = 0 + 35 x 35

v = 1225 m/s

height of the rocket where fuel is empty

v² = u² + 2 a s

1225² = 0 + 2 x 35 x h₁

h₁ = 21437.5 m

After 35 s the rocket will be moving upward till the final velocity becomes zero.

Now, using equation of motion to find the height after 35 s

v² = u² + 2 g h₂

0² = 1225² + 2 x (-9.8) h₂

h₂ = 76562.5 m

total height = h₁ + h₂

          = 76562.5 m + 21437.5 m = 98000 m

now, time taken by before the rocket hit the ground

using equation of motion

s = u t +\dfrac{1}{2}at^2

-13500 = 1225 t -\dfrac{1}{2}\times 9.8 \times t^2

negative sign is used because the distance travel by the rocket is downward.

4.9 t² - 1225 t - 13500 = 0

t = \dfrac{-(-1225)\pm \sqrt{1225^2 - 4\times 4.9 \times (-13500)}}{2\times 4.9}

t = 260.57 s

neglecting the negative sign

total time the rocket was in air

T = t₁ + t₂

T = 35 + 260.57

T = 295.57 s

Time for which rocket was in air is equal to 295.57 s.

6 0
3 years ago
СРОЧНО ПОМОГИТЕ ПОЖАЛУЙСТА!!!!!!!!!!<br>​
Deffense [45]

Боже, как это сложно! Ну ладно.

Между прочим это ты сам должен делать, а то не куда не поступишь!

3 0
3 years ago
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