1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
DedPeter [7]
2 years ago
14

The go kart driver does 1234J of work when hitting a barrier. Assuming the barrier moved 100 meters from its original location,

what was the force at which the driver hit the barrier?
Physics
2 answers:
leva [86]2 years ago
6 0
Force=1234J*100m
So the answer is 123400

bezimeni [28]2 years ago
4 0
The guy in front of me is correct
You might be interested in
A 1.0 ball moving at 2.0 / perpendicular to a wall rebounds from the wall at 1.5 /. If the ball was in contact with the wall for
xxTIMURxx [149]

Answer:

5N

Explanation:

We have a simple problem of momentum here.

ΔMomentum= mΔv= FΔt

Solve for F

mΔv/Δt=F

Plug in givens

1*(2-1.5)/0.1=F

F=5N

4 0
2 years ago
At what time was the person at a position of 0m?
Anni [7]

Answer: The person was not at a position of "0" at any time. The person started at 10 metres from the starting line. The explanation below shows how to use the standard formula for position when the initial position is not "0". It is noteworthy that the standard expression of the formula for distance travelled does not include a variable (e.g. "d") for distance at the start (when t(time) = 0)

Explanation: At time = 0, the start, the person was at 10m distance from the starting line. Therefore, to use the standard equation, "s + ut + 1/2att (t squared, that is), distance from starting line = 10 + s, that is, total distance from starting line  equals initial position, 10 metres, plus "s" (distance travelled from t = 0 to t = 1) in metres.

for the section of the graph from "0" seconds (t = 0) to 1 second (t = 1):

s = ut + 1/2att

the initial position is 10 metres.

s = 10

the distance is constant from t = 0 to t = 1, therefore the velocity for the whole of that section of graph must be 0.

u = 0

there is no change in the velocity from t = 0 to t= 1, therefore the acceleration for the first section of the graph must be 0.

a = 0

s = ut + 1/2att

  = (0 x 1) + 1/2 (0 x 1 x 1)

  = (0) + 1/2 (0)

  = 0

total distance from starting line (position) equals initial position plus change in position (distance travelled).

at t = 1,

position = 10 + 0

 = 10 metres

The whole of the graph can be analysed using this process for each straight section of the graph separately, adding "s" for each section to the previous total of distance from starting line.

using "d" for initial distance from starting line ( position ), d1 for distance from starting line at t = 1, d2 for distance from starting line at t = 2, etcetera:

section 1, t = 0 to t = 1:

d1 (t=0 to t=1)  =  10 + s (t=0 to t=1).

section 2, t= 1 to t = 2:

d2 (t=0 to t=2) = 10 + s (t=0 to t=1) + s (t=1 to t=2).

etcetera.

8 0
3 years ago
Differentiate the following functions with respect to x <br>xsin x​
gtnhenbr [62]

Answer:

Here is your answer

Hope it helps

6 0
3 years ago
Read 2 more answers
A particle executes simple harmonic motion with an amplitude of 1.69 cm. At what positive displacement from the midpoint of its
m_a_m_a [10]

Answer: 0.0146m

Explanation: The formula that defines the velocity of a simple harmonic motion is given as

v = ω√A² - x²

Where v = linear velocity, A = amplitude = 1.69cm = 0.0169m, x = displacement.

The maximum speed of a simple harmonic motion is derived when x = A, hence v = ωA

One half of maximum speed = speed of motion

3ωA/2 = ω√A² - x²

ω cancels out on both sides of the equation, hence we have that

A/2 = √A² - x²

(0.0169)/2 = √(0.0169² - x²)

0.00845 = √(0.0169² - x²)

By squaring both sides, we have that

0.00845² = 0.0169² - x²

x² = 0.0169² - 0.00845²

x² = 0.0002142

x = √0.0002142

x = 0.0146m

5 0
3 years ago
PLEASE HELP !!
kumpel [21]

Answer:

Movement Time

explanation:

5 0
2 years ago
Other questions:
  • When two waves meet they combine according to the ____________________ principle?
    14·1 answer
  • PHYSICS *30 POINTS + BRAINLIEST ANSWER*
    14·1 answer
  • A 2.00-kg object is attached to a spring and placed on a frictionless, horizontal surface. A horizontal force of 20.0 N is requi
    5·1 answer
  • How much energy is needed to heat up 1kg of water by 15degrees?
    8·1 answer
  • In explaining other people’s behaviors people from western cultures tend to emphasize blank factors and people from eastern cult
    10·1 answer
  • What statement is true because of Newton's Second Law?
    10·2 answers
  • QUESTION 6
    15·2 answers
  • An unusual lightning strike has a vertical portion with a current of –400 A downwards. The Earth’s magnetic field at that locati
    15·1 answer
  • Think about a situation at school where you or someone you know
    10·1 answer
  • ______ is a simple movement in which two opposing surfaces slide slightly back-and-forth or side-to-side with respect to one ano
    15·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!