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Solnce55 [7]
3 years ago
10

How can a driver best prepare to enter sharp curves in the roadway?

Physics
1 answer:
gogolik [260]3 years ago
4 0
They should practice using their brakes on the car so they can get better at driving around sharp curves
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You might be interested in
Can anyone help me with this question please​
JulsSmile [24]

Explanation:

V=u+at

where,

v=final speed

u=initial speed,(starting speed)

a=acceleration

t=time

  1. v=u+at = 6=2+a*2

6=2+2a

2a=6-2

2a=4

a=4/2 = 2

a =2

2. to find time taken

v=u+at

25=5*2t

2t=25-5

2t=20

t=20/2

t=10sec

3. finding final speed

v=u+at

v=4+10*2

=4+20

v=24m/sec

5.v=u+at

=5+8*10

=5+80

V=85m/sev

6. v=u+at

8=u+4*2

8=u+8

U=8/8

u=1

these are your missing values

5 0
3 years ago
Please help! This is due tomorrow and I absolutely need help.
zysi [14]

Answer:

Correct answer:  11. Total distance d = 200m ; 12. Vav = 3.63m/s ;

13. Total displacement Dt = 0m ; 14. V₂(10s-15s) = 0 m/s ;

15. V₃(15s-40s) = 4 m/s ; 16. V₁(0s-10s) = 6 m/s > V₄(40s-55s) = 2.67 m/s

Explanation:

The whole movement can be divided into four stages.

In the first stage the subject moves 60m in a positive direction for 10s,

in the other it is stationary for 5s, in the third it moves 100m in the opposite (negative) direction for 25s and in the fourth in the positive 40m for 15s.

11. Total distance = 60 + 0 + 100 + 40 = 200m

12. The formula for calculating the average speed (velocity) is

Vav = (S₁ + S₂ + S₃ + S₄) / (t₁ + t₂ + t₃ + t₄)

Vav = (60 + 0 + 100 + 40)/ (10 + 5 + 25 + 15) = 200/55 = 3.63 m/s

13. The movement started from the origin and ended at the origin

Total displacement is zero meters.

14. The speed between 10s and 15s is zero, because he did not move.

15. V₃ = S₃/t₃ = 100/25 = 4 m/s

16. V₁ = S₁/t₁ = 60/10 = 6 m/s   and V₄ = S₄/t₄ = 40/15 = 2.67 m/s

V₁ > V₄

God is with you!!!

4 0
3 years ago
A uniform disk is constrained to rotate about an axis passing through its center and perpendicular to the plane of the disk. If
ella [17]

Answer:

442.5 rad

Explanation:

w₀ = initial angular velocity of the disk = 7.0 rad/s

α = Constant angular acceleration = 3.0 rad/s²

t = time period of rotation of the disk = 15 s

θ = angular displacement of the point on the rim

Angular displacement of the point on the rim is given as

θ = w₀ t + (0.5) α t²

inserting the values

θ = (7.0) (15) + (0.5) (3.0) (15)²

θ = 442.5 rad

4 0
2 years ago
Raw sewage is raised vertically by 5.49m in the amountof
uranmaximum [27]

Answer:

          P = 1235.7646 W

Explanation:

Given data:

height of raised sewage = 5.49 m

rate of sewage =1.89*10^6 lt/day

density of sewage = 1.050 kg/m^3

power is written as

P = \frac{work}{time}

work  = m g h

m - mass = ( \rho V)

  h -height

mass lifted per day = ( 1.89*10^6)(1.050)

                              = 1984500 kg

time = 24 hours* 3600 seconds per hour

powerP = \frac{( 1984500)(9.8)( 5.49)}{(24)(3600)}

          P = 1235.7646 W

5 0
3 years ago
You accelerate a 0.43kg football 5m/s2. Calculate the force you applied to the football.
Elza [17]

The force applied on the football is 2.15 Newton.

Given the data in the question;

  • Mass of football; m = 0.43kg
  • Acceleration; a = 5m/s^2

Force applied; F = \ ?

To determine the force applied on the football, we use Newton's laws of motion:

F = m * a

Where m is the mass of the object and a is the acceleration.

We substitute our given values into the equation

F = 0.43kg\ *\ 5m/s^2\\\\F = 2.15kg.m/s^2\\\\F = 2.15N

Therefore, the force applied on the football is 2.15 Newton.

Learn more: brainly.com/question/2388393

7 0
2 years ago
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