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OLga [1]
3 years ago
12

How might an intense solar storm affect people on Earth?

Physics
1 answer:
kow [346]3 years ago
8 0

An intense solar storm COULD disrupt communications and damage the power grid. <em> (A)</em>

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A car of mass 1000 kg is travelling at 10 m/s. What is its kinetic energy?
Marysya12 [62]

Answer:

50,000

Explanation:

kinetic energy=1\2×mv^2

=1\2×1000kg×10×10

=50000

6 0
2 years ago
A student is constructing an investigation on static electricity. The student has three balloons and rubs two of them on a piece
Kamila [148]

Answer:

Explanation:

As given, the student has three balloons and rubs two of them on a piece of wool. The rubbing of balloon on wool is the independent variable as it was done on two and not on the third as control.

5 0
2 years ago
Charging a balloon by rubbing it on wool is an example of
SVETLANKA909090 [29]
Electronic friction ^.^
3 0
3 years ago
Read 2 more answers
Two particles, each of mass 7.0 kg, are a distance 3.0 m apart. To bring a third particle, with mass 21 kg, from far away to a r
garri49 [273]

Answer: the external agent must do work equal to -1.3 × 10⁻⁸ J

Explanation:

Given that;

Mass M1 = 7.0 kg

r = 3.0/2 m = 1.5 m

Mass M2 = 21 kg

we know that G = 6.67 × 10⁻¹¹ N.m²/kg²

work done by an external agent W = -2GM2M1 / r

so we substitute

W = (-2 × 6.67 × 10⁻¹¹ × 21 × 7) / 1.5

W = -1.96098 × 10⁻⁸ / 1.5

W = -1.3 × 10⁻⁸ J

Therefore the external agent must do work equal to -1.3 × 10⁻⁸ J

8 0
3 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
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