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larisa86 [58]
3 years ago
9

Observing a spacecraft land on a distant asteroid, scientists notice that the craft is falling at a rate of 5 m/s. When it is 10

0 m closer to the surface of the asteroid, the craft reports a velocity of 8 m/s. According to their data, what is the approximate gravitational acceleration on this asteroid?
A. 0 m/s2
B. 0.03 m/s2
C. 0.20 m/s2
D. 0.65 m/s2
E. 33 m/s2
Physics
1 answer:
denpristay [2]3 years ago
4 0

Answer:

(c) 0.20 m/s²

Explanation:

Using The equation of motion,

V² = u² + 2gs...................... equation 1

Where V = final velocity, u = initial velocity, g = gravitational acceleration,

s = distance.

Making g the subject of the equation in equation 1 above,

g =( V² - u²)/2s............................ equation 2

Where V= 8 m/s, u = 5 m/s, s= 100 m.

g = (8²-5²)/(2×100)

g = (64-25)/200

g = 39/200

g =0.195 m/s²

g ≈ 0.20 m/s².

The gravitational acceleration = 0.20 m/s²

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When the temperature of the air is 50°C, the velocity of a sound wave traveling through the air is approximately?
noname [10]

The answer is:

C. 361 m/s

The explanation:

To calculate  the speed of sound at a given temperature (50°C) we are going to use this formula:

v = 331 + 0.6T

when V is the velocity

and T is the temperature = 50°C

by substitution:

v = 331 + 0.6(50)

v = 361 m/s

So, The correct answer is C.

because of the variation of the motion of the molecules of air with change of temperature so, the velocity (V) of the  sound in the air is change with temperature.

5 0
4 years ago
Read 2 more answers
Find electric field at point p which is a distance l away from the both +q and -q
denis-greek [22]

Answer:

\frac{1}{4\times(pie)\times\text{E}} \times\frac{q}{I^{2} }+\frac{1}{4\times(pie)\times\text{E}} \times\frac{-q}{I^{2} }

Explanation:

As given point p is equidistant from both the charges

It must be in the middle of both the charges

Assuming all 3 points lie on the same line

Electric Field due a charge q at a point ,distance r away

=\frac{1}{4\times(pie)\times\text{E}} \times\frac{q}{r^{2} }

Where

  • q is the charge
  • r is the distance
  • E is the permittivity of medium

Let electric field due to charge q be F1 and -q be F2

I is the distance of P from q and also from charge -q

⇒

F1=\frac{1}{4\times(pie)\times\text{E}} \times\frac{q}{I^{2} }

F2=\frac{1}{4\times(pie)\times\text{E}} \times\frac{-q}{I^{2} }

⇒

F1+F2=\frac{1}{4\times(pie)\times\text{E}} \times\frac{q}{I^{2} }+\frac{1}{4\times(pie)\times\text{E}} \times\frac{-q}{I^{2} }

8 0
3 years ago
Why does convection occur only in fluids?
irinina [24]
Because the particles of a solid are not free to move,
no matter how hot and buoyant they become.
3 0
4 years ago
After being struck by a bowling ball, a 1.3 kg bowling pin sliding to the right at 5.0 m/s collides head-on with another 1.3 kg
GuDViN [60]

Answer:

a) 4.2m/s

b) 5.0m/s

Explanation:

This problem is solved using the principle of conservation of linear momentum which states that in a closed system of colliding bodies, the sum of the total momenta before collision is equal to the sum of the total momenta after collision.

The problem is also an illustration of elastic collision where there is no loss in kinetic energy.

Equation (1) is a mathematical representation of the the principle of conservation of linear momentum for two colliding bodies of masses m_1 and m_2 whose respective velocities before collision are u_1 and u_2;

m_1u_1+m_2u_2=m_1v_1+m_2v_2..............(1)

where v_1 and v_2 are their respective velocities after collision.

Given;

m_1=1.3kg\\u_1=5m/s\\m_2=1.3kg\\u_2=0m/s

Note that u_2=0 because the second mass m_2 was at rest before the collision.

Also, since the two masses are equal, we can say that m_1=m_2=m so that equation (1) is reduced as follows;

mu_1+mu_2=mv_1+mv_2\\\\m(u_1+u_2)=m(v_1+v_2)..............(2)

m cancels out of both sides of equation (2), and we obtain the following;

u_1+u_2=v_1+v_2.............(3)

a) When v_1=0.8m/s, we obtain the following by equation(3)

5+0=0.8+v_2\\hence\\v_2=5-0.8\\v_2=4.2m/s

b) As m_1 stops moving v_1=0, therefore,

5+0=0+v_2\\v_2=5m/s

5 0
3 years ago
A 4kg object is moving at a speed of 5 m/sec.how much kinetic energy does the objects have
nikklg [1K]

Answer:

20

Explanation:

3 0
3 years ago
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