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nadezda [96]
3 years ago
6

An asteroid is speeding directly towards a space ship with a velocity of 255 m/s. If the asteroid is detected 8000 meters from t

he space ship, how long does the crew have to take evasive action before the asteroid hits the ship?
A) about 22.6 seconds
B) about 31.4 seconds
C) about 32.0 seconds
D) about 41.5 seconds
Physics
2 answers:
Softa [21]3 years ago
5 0
B) about 31.4 seconds
Sveta_85 [38]3 years ago
3 0
B) 31.4 I got this by doing 8,000 divided by 255.




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Elza [17]
In this item, we are asked to determine the speed of the bobsled given the distance traveled and the time it takes to cover the certain distance. This can mathematically be expressed as,
                          speed = distance / time

Substituting the given values in this item,
                         speed = (113 m) / (29 s)
                         speed = 3.90 m/s

<em>ANSWER: 3.90 m/s</em>
3 0
3 years ago
If the pressure exerted on a 300.0 mL sample of hydrogen gas at constant temperature is increased from 0.500 kPa to 0.750 kPa, w
uranmaximum [27]

Answer:

200 mL

Explanation:

Given that,

Initial volume, V₁ = 300 mL

Initial pressure, P₁ = 0.5 kPa

Final pressure, P₂ = 0.75 kPa

We need to find the final volume of the sample if pressure is increased at constant temperature. It is based on Boyle's law. Its mathematical form is given by :

V\propto \dfrac{1}{P}\\\\P_1V_1=P_2V_2

V₂ is the final volume

V_2=\dfrac{P_1V_1}{P_2}\\\\V_2=\dfrac{300\times 0.5}{0.75}\\\\V_2=200\ mL

So, the final volume of the sample is 200 mL.

5 0
2 years ago
How was Bohr's atomic model different from those of previous scientists?
exis [7]

Answer:Bohr placed the electrons in distinct energy levels.  Rutherford described the atom as consisting of a tiny positive mass surrounded by a cloud of negative electrons. Bohr thought that electrons orbited the nucleus in quantised orbits.

Explanation: also rutherfords was just a hypothesis while Bhor took the time to make his an experiment

4 0
3 years ago
A student at another university repeats the experiment you did in lab. Her target ball is 0.860 m above the floor when it is in
dexar [7]

Answer:

K = 0.076 J

Explanation:

The height of the target, h = 0.860  m

The mass of the steel ball, m = 0.0120 kg

Distance moved, d = 1.50 m

We need to find the kinetic energy (in joules) of the target ball just after it is struck. Let t is the time taken by the ball to reach the ground.

h=ut+\dfrac{1}{2}at^2\\\\t=\sqrt{\dfrac{2h}{g}}

Put all the values,

t=\sqrt{\dfrac{2\times 0.860 }{9.8}} \\\\=0.418\ s

The velocity of the ball is :

v=\dfrac{1.5}{0.418}\\\\= $$3.58\ m/s

The kinetic energy of the ball is :

K=\dfrac{1}{2}mv^2\\\\K=\dfrac{1}{2}\times 0.0120\times 3.58^2\\\\=0.076\ J

So, the required kinetic energy is 0.076 J.

6 0
3 years ago
The sun emits electromagnetic waves with a power of 4.0 × 10²⁶ W. Determine the intensity of electromagnetic waves from the sun
Sphinxa [80]

Answer:

I_v = 2,700 W / m^2

I_m = 610 W / m^2

I_s = 16 W / m^2

Explanation:

Given:

- The Power of EM waves emitted by Sun P_s = 4.0*10^26 W

- Radius of Venus r_v = 1.08 * 10^11 m

- Radius of Mars r_m = 2.28 * 10^11 m

- Radius of Saturn r_s = 1.43 * 10^12 m

Find:

Determine the intensity of electromagnetic waves from the sun just outside the atmospheres of (a) Venus, (b) Mars, and (c) Saturn.

Solution:

- We know that Power is related to intensity and surface area of an object follows:

                                        I = P / 4*pi*r^2

Where, A is the surface area of a sphere models the atmosphere around the planets.

a)

- The intensity at the surface of Venus is calculated as:

                                       I_v = P_s / 4*pi*r^2_v

                                       I_v = 4.0*10^26 / 4*pi*(1.08*10^11)^2

                                       I_v = 2,700 W / m^2

b)

- The intensity at the surface of Mars is calculated as:

                                       I_m = P_s / 4*pi*r^2_m

                                       I_m = 4.0*10^26 / 4*pi*(2.28*10^11)^2

                                      I_m = 610 W / m^2

c)

- The intensity at the surface of Saturn is calculated as:

                                       I_s = P_s / 4*pi*r^2_s

                                       I_s = 4.0*10^26 / 4*pi*(1.43*10^12)^2

                                      I_s = 16 W / m^2

7 0
3 years ago
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