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Novay_Z [31]
1 year ago
10

On the earth, when an astronaut throws a 0.250-kg stone vertically upward, it returns to his hand a time T later. On planet X he

finds that, under the same circumstances, the stone returns to his hand in 2T. In both cases, he throws the stone with the same initial velocity and it feels negligible air resistance. The acceleration due to gravity on planet X (in terms of g) is _______.a) g/. b) g/4. c) 2g. d) g/2. e) g.
Physics
1 answer:
Liula [17]1 year ago
4 0

Answer:

correct is d) a ’= g / 2

Explanation:

For this exercise let's use the kinematics equations

On earth

      v = v₀ - a t

     a = (v₀- v) / T

On planet X

    v = v₀ - a' t’

    a ’= (v₀-v) / 2T

Let's substitute the land values ​​in plot X

     a’= a / 2

Now let's use Newton's second law

       W = ma

      m g = m a

      a = g

We substitute

      a ’= g / 2

So we see that on planet X the acceleration is half the acceleration of Earth's gravity

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A spaceship of frontal area 10 m2 moves through a large dust cloud with a speed of 1 x 106 m/s. The mass density of the dust is
Step2247 [10]

Answer:

The decelerating force is 3\times 10^{- 11}\ N

Solution:

As per the question:

Frontal Area, A = 10\ m^{2}

Speed of the spaceship, v = 1\times 10^{6}\ m/s

Mass density of dust, \rho_{d} = 3\times 10^{- 18}\ kg/m^{3}

Now, to calculate the average decelerating force exerted by the particle:

Mass,\ m = \rho_{d}V                                (1)

Volume, V = A\times v\times t

Thus substituting the value of volume, V in eqn (1):

m = \rho_{d}(Avt)

where

A = Area

v = velocity

t = time

m = \rho_{d}(A\times v\times t)                  (2)

Momentum,\ p = \rho_{d}(Avt)v = \rho_{d}Av^{2}t

From Newton's second law of motion:

F = \frac{dp}{dt}

Thus differentiating w.r.t time 't':

F_{avg} = \frac{d}{dt}(\rho_{d}Av^{2}t) = \rho_{d}Av^{2}

where

F_{avg} = average decelerating force of the particle

Now, substituting suitable values in the above eqn:

F_{avg} = 3\times 10^{- 18}\times 10\times 1\times 10^{6} = 3\times 10^{- 11}\ N

4 0
1 year ago
In a harbor, you can see sea waves traveling around the edges of small stationary boats. Why does this happen?
faust18 [17]
Below are the choices that can be found in the other sources:

A. diffraction 
<span>B. refraction </span>
<span>C. reflection </span>
<span>D. transmission
</span>
The answer is diffraction. It means that <span>the process by which a beam of light or other system of waves is spread out as a result of passing through a narrow aperture or across an edge, typically accompanied by interference between the wave forms produced.</span>
8 0
1 year ago
Sayid made a chart listing data of two colliding objects. A 5-column table titled Collision: Two Objects Stick Together with 2 r
Alborosie

Answer:

6 m/s is the missing final velocity

Explanation:

From the data table we extract that there were two objects (X and Y) that underwent an inelastic collision, moving together after the collision as a new object with mass equal the addition of the two original masses, and a new velocity which is the unknown in the problem).

Object X had a mass of 300 kg, while object Y had a mass of 100 kg.

Object's X initial velocity was positive (let's imagine it on a horizontal axis pointing to the right) of 10 m/s. Object Y had a negative velocity (imagine it as pointing to the left on the horizontal axis) of -6 m/s.

We can solve for the unknown, using conservation of momentum in the collision: Initial total momentum = Final total momentum (where momentum is defined as the product of the mass of the object times its velocity.

In numbers, and calling P_{xi} the initial momentum of object X and P_{yi} the initial momentum of object Y, we can derive the total initial momentum of the system: P_{total}_i=P_{xi}+P_{yi}= 300*10 \frac{kg*m}{s} -100*6\frac{kg*m}{s} =\\=(3000-600 )\frac{kg*m}{s} =2400 \frac{kg*m}{s}

Since in the collision there is conservation of the total momentum, this initial quantity should equal the quantity for the final mometum of the stack together system (that has a total mass of 400 kg):

Final momentum of the system: M * v_f=400kg * v_f

We then set the equality of the momenta (total initial equals final) and proceed to solve the equation for the unknown(final velocity of the system):

2400 \frac{kg*m}{s} =400kg*v_f\\\frac{2400}{400} \frac{m}{s} =v_f\\v_f=6 \frac{m}{s}

7 0
2 years ago
Read 2 more answers
(YOU WILL GET BRAINLIEST)Matter may be classified as a pure substance or a mixture. Where on the Venn diagram would you insert t
inessss [21]
I think it would be B because it is matter, since it has atoms, and it contains subatomic particles, which are smaller than atoms
3 0
1 year ago
Read 2 more answers
A steel sphere sits on top of an aluminum ring. The steel sphere (a= 1.1 x 10^-5/degrees celsius) has a diameter of 4.000 cm at
mote1985 [20]

Answer:

C

Explanation:

To solve this question, we will need to develop an expression that relates the diameter 'd', at temperature T equals the original diameter d₀ (at 0 degrees) plus the change in diameter from the temperature increase ( ΔT = T):

d = d₀ + d₀αT

for the sphere, we were given

D₀ = 4.000 cm

α = 1.1 x 10⁻⁵/degrees celsius

we have D = 4 + (4x(1.1 x 10⁻⁵)T = 4 + (4.4x10⁻⁵)T             EQN 1

Similarly for the Aluminium ring we have

we were given

d₀ = 3.994 cm

α = 2.4 x 10⁻⁵/degrees celsius

we have d = 3.994 + (3.994x(2.4 x 10⁻⁵)T = 3.994 + (9.58x10⁻⁵)T       EQN 2

Since @ the temperature T at which the sphere fall through the ring, d=D

Eqn 1 = Eqn 2

4 + (4.4x10⁻⁵)T =3.994 + (9.58x10⁻⁵)T, collect like terms

0.006=5.18x10⁻⁵T

T=115.7K

8 0
1 year ago
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