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Nezavi [6.7K]
3 years ago
8

The roller coaster is designed according to safety regulations that prohibit the speed of the car from exceeding 20 m/s. find th

e maximum value of a allowed by these regulations.
Physics
1 answer:
hram777 [196]3 years ago
4 0

maximum allowed value of the speed in roller coaster is given as

v = 20 m/s

now from kinematics we can say

v^2 - v_i^2 = 2 a s

here initial speed will be

v_i = 0

acceleration is due to gravity

a = 9.8 m/s^2

now we can use this to find the height

20^2 - 0^2 = 2 * 9.8* h

400 = 19.6 *h

h = 20.4 m

so maximum allowed height will be 20.4 m

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in the figure shown, if the mass of the block is 4kg and the coefficient of static friction is 0.5 and the coefficient of kineti
Elan Coil [88]

Answer:

Ff = 19.6 N

Explanation:

So since its saying whats the minimum F to move the block, we will use static friction (0.5).

We will use the equation for force of friction, which is Ff = uFn

Ff = (0.5)(4)(9.8)

Ff = 19.6 N

this is the minumum force needed to move the block, as that is the frictional force. You would need to apply a minimum force of 19.6 N to move the block

3 0
3 years ago
WILL GIVE BRAINLIEST TO CORRECT ANSWER PLEASE HELP ME
koban [17]

Answer:

The total distance is 381.5 [m]

Explanation:

In order to solve this problem we must use the expressions of kinematics. The clue to solve this problem is that the motorcyclist starts from rest, i.e. its initial speed is zero.

v_{f} =v_{o} +(a*t)

where:

Vf = final velocity [m/s]

Vo = initial velocity = 0

a = acceleration = 2 [m/s²]

t = time = 7 [s]

Vf = 0 + (2*7)

Vf = 14 [m/s]

With this velocity, we can calculate the displacement using the following expression.

v_{f} ^{2} =v_{o} ^{2} +2*a*x

where

x = distance traveled [m]

14² = 0 + (2*7*x)

x = 196/(14)

x = 14 [m]

Note: The positive sign in the equations is because the car is accelerating, it means its velocity is increasing.

The other important clue to solve this problem in the second part is that the final velocity is now the initial velocity.

We must calculate the final velocity.

v_{f}= v_{i} +(a*t)

Vf = final velocity [m/s]

Vi = initial velocity = 14 [m/s]

a = desacceleration = 4 [m/s²]

t = time = 8 [s]

Vf = 24 + (4*8)

Vf = 56 [m/s]

With this velocity, we can calculate the displacement using the following expression.

v_{f} ^{2} =v_{o} ^{2} +2*a*x

where

x = distance traveled [m]

56² = 14² + (2*4*x)

x = 2940/(8)

x = 367.5 [m]

Note: The positive sign in the equations is because the car is accelerating, it means its velocity is increasing.

Therefore the total distance is Xt = 14 + 367.5 = 381.5 [m].

4 0
3 years ago
Choose the correct statement regarding the sign conventions for lenses.
Savatey [412]

Answer:

a) false

b) True

c) True

d) False

e) False

Explanation:

a) False

For a diverging lens, the focal length is negative while it is positive for a converging lens

b) True

Image distances  for virtual images are always negative and it also forms on the the same side of the lens as the object and is enlarged

c) True

d) False

For a diverging lens, the focal length is negative while it is positive for a converging lens

e) False

Image distances  for virtual images are always negative and it also forms on the the same side of the lens as the object and is enlarged

3 0
3 years ago
And event or situation that causes a strain on your body or mind is called a help asap
faltersainse [42]
I'm quite certain the answer is "stress".
7 0
3 years ago
An archer defending a castle is on an 15.5 m high wall. He shoots an arrow straight down at 22.8 m/s. How much time does it take
kodGreya [7K]

Answer:

  about 602 milliseconds

Explanation:

The motion can be approximated by the equation ...

  y = -4.9t^2 -22.8t +15.5

where t is the time since the arrow was released, and y is the distance above the ground.

When y=0, the arrow has hit the ground.

Using the quadratic formula, we find ...

  t = (-(-22.8) ± √((-22.8)^2 -4(-4.9)(15.5)))/(2(-4.9))

  = (22.8 ± √823.64)/(-9.8)

The positive solution is ...

  t ≈ 0.60195193

It takes about 602 milliseconds for the arrow to reach the ground.

8 0
3 years ago
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