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saul85 [17]
3 years ago
15

What event was seen as a symbol of the gradual relationship improvements between the United States and the Soviet Union as the C

old War was coming to an end?
A. The Soviet space center and the US space center signed a treaty agreeing to put the past behind them and work together for future space endeavors.
B. A handshake that occurred between the commanders of a US and Soviet space craft that were involved in a joint moon mission.
C. Astronauts from both countries taking a trip to the moon together aboard one space craft.
D. A male astronaut from the Soviet Union and a female astronaut from the US walked on the moon together.
Physics
1 answer:
Olegator [25]3 years ago
4 0

Answer:

I think the answer is b but i cannot confirm cuz this is a challenging question

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When a 0.622 kg basketball hits the floor, its velocity changes from 4.23 m/s down to 3.85 m/s up. If the ball was in contact wi
SVEN [57.7K]

when the ball hits the floor and bounces back the momentum of the ball changes.

the rate of change of momentum is the force exerted by the floor on it.

the equation for the force exerted is

f = rate of change of momentum

f = \frac{mv - mu}{t}

v is the final velocity which is - 3.85 m/s

u is initial velocity - 4.23 m/s

m = 0.622 kg

time is the impact time of the ball in contact with the floor - 0.0266 s

substituting the values

f = \frac{0.622 kg (3.85 m/s - (-)4.23 m/s)}{0.0266}

since the ball is going down, we take that as negative and ball going upwards as positive.

f = 189 N

the force exerted from the floor is 189 N

4 0
3 years ago
Solar panels convert light energy from sunlight into electrical energy. What material is most likely used in solar panels, and w
klemol [59]

Answer:

A metalloid is used because it is a semiconductor and can become more conductive when more light shines on it

Explanation:

The material used in a solar panel is a metalloid. It can often become conductive when more light shines on it.

Metalloids have properties that straddles between those of metals and non-metals.

In essence, they can be conductive or not under certain conditions.

The most important property they exhibit is that they can become more conductive when more light shines on them. This way more electrons are produced.

3 0
3 years ago
Read 2 more answers
A person can jump 1.5m on the earth. How high could the person jump on a planet having the twice the mass of the earth and twice
MrMuchimi
F=mg=Gm1m2/r^2
g=Gm2/r^2
g=2Gm2/(2r)^2=2Gm2/4r^2=Gm2/2r^2
So since there is half times the gravity on this unknown planet that has twice earth's mass and twice it's radius, then the person can jump twice as high. 1.5*2= 3m high

5 0
4 years ago
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A uniform disk with mass 35.2 kg and radius 0.200 m is pivoted at its center about a horizontal, frictionless axle that is stati
Sergio [31]

Answer:

a) v = 1.01 m/s

b) a = 5.6 m/s²

Explanation:

a)

  • If the disk is initially at rest, and it is applied a constant force tangential to the rim, we can apply the following expression (that resembles Newton's 2nd law, applying to rigid bodies instead of point masses) as follows:

       \tau = I * \alpha  (1)

  • Where τ is the external torque applied to the body, I is the rotational inertia of the body regarding the axis of rotation, and α is the angular acceleration as a consequence of the torque.
  • Since the force is applied tangentially to the rim of the disk, it's perpendicular to the radius, so the torque can be calculated simply as follows:
  • τ = F*r (2)
  • For a solid uniform disk, the rotational inertia regarding an axle passing through its center  is just I = m*r²/2 (3).
  • Replacing (2) and (3) in (1), we can solve for α, as follows:

       \alpha = \frac{2*F}{m*r} = \frac{2*34.5N}{35.2kg*0.2m} = 9.8 rad/s2 (4)

  • Since the angular acceleration is constant, we can use the following kinematic equation:

        \omega_{f}^{2}  - \omega_{o}^{2} = 2*\Delta \theta * \alpha (5)

  • Prior to solve it, we need to convert the angle rotated from revs to radians, as follows:

       0.2 rev*\frac{2*\pi rad}{1 rev} = 1.3 rad (6)

  • Replacing (6) in (5), taking into account that ω₀ = 0 (due to the disk starts from rest), we can solve for ωf, as follows:

       \omega_{f} = \sqrt{2*\alpha *\Delta\theta} = \sqrt{2*1.3rad*9.8rad/s2} = 5.1 rad/sec (7)

  • Now, we know that there exists a fixed relationship the tangential speed and the angular speed, as follows:

        v = \omega * r (8)

  • where r is the radius of the circular movement. If we want to know the tangential speed of a point located on the rim of  the disk, r becomes the radius of the disk, 0.200 m.
  • Replacing this value and (7) in (8), we get:

       v= 5.1 rad/sec* 0.2 m = 1.01 m/s (9)

b)    

  • There exists a fixed relationship between the tangential and the angular acceleration in a circular movement, as follows:

       a_{t} = \alpha * r (9)

  • where r is the radius of the circular movement. In this case the point is located on the rim of the disk, so r becomes the radius of the disk.
  • Replacing this value and (4), in (9), we get:

       a_{t}  = 9.8 rad/s2 * 0.200 m = 1.96 m/s2 (10)

  • Now, the resultant acceleration of a point of the rim, in magnitude, is the vector sum of the tangential acceleration and the radial acceleration.
  • The radial acceleration is just the centripetal acceleration, that can be expressed as follows:

       a_{c} = \omega^{2} * r  (11)

  • Since we are asked to get the acceleration after the disk has rotated 0.2 rev, and we have just got the value of the angular speed after rotating this same angle, we can replace (7) in (11).
  • Since the point is located on the rim of the disk, r becomes simply the radius of the disk,, 0.200 m.
  • Replacing this value and (7) in (11) we get:

       a_{c} = \omega^{2} * r   = (5.1 rad/sec)^{2} * 0.200 m = 5.2 m/s2 (12)

  • The magnitude of the resultant acceleration will be simply the vector sum of the tangential and the radial acceleration.
  • Since both are perpendicular each other, we can find the resultant acceleration applying the Pythagorean Theorem to both perpendicular components, as follows:

       a = \sqrt{a_{t} ^{2} + a_{c} ^{2} } = \sqrt{(1.96m/s2)^{2} +(5.2m/s2)^{2} } = 5.6 m/s2 (13)

6 0
3 years ago
If a metal has a mass of 78.2 g and a volume of 11.4 mL, what is its density?
hram777 [196]

Answer:

6.86g/mL

Explanation:

4 0
3 years ago
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