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storchak [24]
2 years ago
12

A car is traveling in a race. The car went from the initial velocity of 35 m/s to the final velocity of 65 m/s in 5 seconds. Wha

t is the acceleration? 0-13 m/s2 0-6 m/s2 0.6 m/s2 O 13 m/s2​
Physics
1 answer:
olga55 [171]2 years ago
3 0

Answer:

6m/s²

Explanation:

Given parameters:

Initial velocity  = 35m/s

Final velocity  = 65m/s

Time taken  = 5s

Unknown:

Acceleration = ?

Solution:

Acceleration is the rate of change of velocity with time taken;

   Acceleration  = \frac{Final velocity  - Initial velocity }{time}  

So;

 Acceleration   = \frac{65  - 35}{5}    = 6m/s²

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zhuklara [117]

Answer:

In a column of fluid, pressure increases with depth as a result of the weight of the overlying fluid. Thus a column of fluid, or an object submerged in the fluid, experiences greater pressure at the bottom of the column than at the top. This difference in pressure results in a net force that tends to accelerate an object upwards.

The pressure at a depth in a fluid of constant density is equal to the pressure of the atmosphere plus the pressure due to the weight of the fluid, or p = p 0 + ρ h g , p = p 0 + ρ h g , 14.4

Granite: 2.70 × 10 32.70 × 10 3

Lead: 1.13 × 10 41.13 × 10 4

Iron: 7.86 × 10 37.86 × 10 3

Oak: 7.10 × 10 27.10 × 10 2

4 0
3 years ago
A camper stands in a valley between two parallel cliff walls. He claps his hands and notices that the echo from the nearby wall
Stells [14]
5,658 ft is your answer I believe
6 0
2 years ago
Define limitations in the operation conditions of a pn junction<br>​
suter [353]

Answer:

Such limitations are given below.

Explanation:

  • Each pn junction provides limited measurements of maximum forwarding current, highest possible inversion voltage as well as the maximum output level.
  • If controlled within certain adsorption conditions, the pn junction could very well offer satisfying performance.
  • In connector operation, the maximum inversion voltage seems to be of significant importance.

6 0
3 years ago
If a 20 g cannonball is shot from a 5 kg cannon with a velocity of 100
balu736 [363]

Strange as it may seem, the statement in the question appears to be <em>TRUE</em>.  

-- Before the shot, neither the cannon nor the ball is moving, so their combined momentum is zero.  

-- Since momentum is conserved, we know immediately that their combined momentum AFTER the shot also has to be zero.

-- (20g is rather puny for a "cannonball" ... about the same weight as four nickels. But we'll take your word for it and just do the Math and the Physics.)

-- Momentum = (mass) x (velocity)

After the shot, the momentum of the cannonball is

(0.02 kg) x (100 m/s ==> that way)

Momentum of the ball = 2 kg-m/s ==> that way.

-- In order for both of them to add up to zero, the momentum of the cannon must be (2 kg-m/s this way <==) .

Momentum of cannon = (5 kg) x (V m/s this way <==)

2 kg-m/s this way <== = (5 kg) x (V m/s this way <==)

Divide each side by (5 kg):

V m/s  =  (2/5) m/s this way <==

Speed of recoil of the cannon = <em>-- 0.4 m/s</em>

3 0
3 years ago
In tae-kwon-do, a hand is slammed down onto a target at a speed of 10 m/s and comes to a stop during the 7.0 ms collision. Assum
wlad13 [49]

Answer:

(A) Impulse = 9Ns

(B) F = 1286N

Explanation:

Impulse = change in momentum = m(v-u)

v = 0 (the hand comes to a stop)

u = -10m/s

Mass = 0.9kg

Impulse = 0.9 ×(0- (-10))

= 9Ns

(B) F×t = Impulse

F = Impulse/ t

t = 7ms = 7×10-³

F = 9/ (7×10-³)

F = 1286N.

5 0
3 years ago
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