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Alenkinab [10]
3 years ago
15

While in empty space, an astronaut throws a ball at a velocity of 11 m/s. What will the velocity of the ball be after it has tra

veled 5 meters?
Physics
2 answers:
iogann1982 [59]3 years ago
7 0

Answer:

Velocity of the ball will be 11 m/sec after traveling 5 meter

Explanation:

We have given that velocity of ball is 11 m/sec in space

We have to find the speed of the ball after it travels 5 meter

We know that in space there is no force acting

And if there is no force in the ball then velocity of the ball will never change, that is the velocity of the ball is always constant

So the velocity of the ball will be 11 m /sec after traveling 11 m/sec

Mashutka [201]3 years ago
6 0
The velocity will be 11 m/s
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Please help!!!! on both of them this is physics not math
Lera25 [3.4K]
Ok this is probably the easiest part in physics so lol dont panic. now the way you round is by looking at the exponent which in this case for the first one is -9 so 6.8 is gonna be 0000000068. because when you have a negative exponent you move the decimal to the west side. when you have a positive decimal you move to the east side. I will not do the second one because its best if you try to solve it so you can learn it. good luck my friend.
3 0
3 years ago
A jet transport with a landing speed of 200 km/h reduces its speed to 60 km/h with a negative thrust R from its jet thrust rever
Amanda [17]

Answer:

257 kN.

Explanation:

So, we are given the following data or parameters or information in the following questions;

=> "A jet transport with a landing speed

= 200 km/h reduces its speed to = 60 km/h with a negative thrust R from its jet thrust reversers"

= > The distance = 425 m along the runway with constant deceleration."

=> "The total mass of the aircraft is 140 Mg with mass center at G. "

We are also give that the "aerodynamic forces on the aircraft are small and may be neglected at lower speed"

Step one: determine the acceleration;

=> Acceleration = 1/ (2 × distance along runway with constant deceleration) × { (landing speed A)^2 - (landing speed B)^2 × 1/(3.6)^2.

=> Acceleration = 1/ (2 × 425) × (200^2 - 60^2) × 1/(3.6)^2 = 3.3 m/s^2.

Thus, "the reaction N under the nose wheel B toward the end of the braking interval and prior to the application of mechanical braking" = The total mass of the aircraft × acceleration × 1.2 = 15N - (9.8 × 2.4 × 140).

= 140 × 3.3× 1.2 = 15N - (9.8 × 2.4 × 140).

= 257 kN.

4 0
3 years ago
Suppose the coefficient of kinetic friction between mA and the plane in the figure(Figure 1) is μk = 0.15, and that mA=mB=2.7kg.
luda_lava [24]
A ) 
T = mB g + mB a
T + mA a - mA g sin 35° = (Mi) mA g cos 35°
------------------------------------------------------------
T = 2.7 · 9.81  + 2.7 a
T = 26.487 + 2.7 a
26.487 + 2.7 a + 2.7 a - 2.7 · 9.81 · 0.574 = 0.15 · 2.7 · 9.81 · 0.819
5.4 a + 26.487 - 15.2023 = 3.2539
5.4 a = 8.0296
a = 1.487 ≈ 1.5 m/s²
B )
T = 2,7 · 9.81 = 26.487 
26.487 - 15.2035 = (Mi) · 2.7 · 9.81 · 0.819
11.2835 = (Mi) · 21.69
(Mi) = 11.2835 : 21.69 = 0.52
4 0
3 years ago
A car driver traveling at a speed of 108km per hour ,sees a traffic light and stopped after travelling for 20seconds .Find the a
navik [9.2K]

Answer:

– 2.5 m/s²

Explanation:

We have,

• Initial velocity, u = 180 km/h = 50 m/s

• Final velocity, v = 0 m/s (it stops)

• Time taken, t = 20 seconds

We have to find acceleration, a.

\longrightarrow a = (v ― u)/t

\longrightarrow a = (0 – 50)/20 m/s²

\longrightarrow a = –50/20 m/s²

\longrightarrow a = – 5/2 m/s²

\longrightarrow a = – 2.5 m/s² (Velocity is decreasing) [Answer]

6 0
3 years ago
Vector a cross b whole square<br><br>​
skad [1K]

Answer:

Kriss go Brrrrrrrrrrr

Explanation:

4 0
3 years ago
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