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tatuchka [14]
3 years ago
11

A car traveling west in a straight line on a highway decreases its speed from 30 meters per second to 23 meters per second in 2

seconds. The car's average acceleration during this time interval is
Physics
2 answers:
garik1379 [7]3 years ago
6 0
U1= 30m/s
u2= 23m/s
t=2s
a=(u2-u1)/t
a=-7/2=-3.5 m/s^2
mihalych1998 [28]3 years ago
3 0
Use the equation Vf=Vi+A(t)
Rearrange it so A is isolated: A= Vf-Vi/t
A=(23-30)/2
A=-3.5m/s^2
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7 0
3 years ago
1. A toy car is marked to go at 0.5m/s, which of the following sets of
Lostsunrise [7]

Answer:

b. 0.6m/s, 0.7m/s, 0.61m/s, 0.62m/s

Explanation:

Precision of a measurement is the closeness of the experimental values to one another. Hence, experimental measurements are said to be precise if they are close to each other irrespective of how close they are to the accepted value. Precision can be determined by finding the range of each experimental value. The measurement with the LOWEST RANGE represents the MOST PRECISE.

Note: Range is the highest value - lowest value

Set A: 1.5 - 0.8 = 0.7

Set B: 0.7 - 0.6 = 0.1

Set C: 2.4 - 2.0 = 0.4

Set D: 3.1 - 2.9 = 0.2

Set B has the lowest range (0.1), hence, represent the most precise value.

8 0
3 years ago
A 2kg block has 70J of KE. It then travels 1.5 meters up a hill. As it travels up the hill friction does -12J of work on the blo
Dima020 [189]

Answer:

v = 5.34[m/s]

Explanation:

In order to solve this problem, we must use the theorem of work and energy conservation. This theorem tells us that the sum of the mechanical energy in the initial state plus the work on or performed by a body must be equal to the mechanical energy in the final state.

Mechanical energy is defined as the sum of energies, kinetic, potential, and elastic.

E₁ = mechanical energy at initial state [J]

E_{1}=E_{pot}+E_{kin}+E_{elas}\\

In the initial state, we only have kinetic energy, potential energy is not had since the reference point is taken below 1.5[m], and the reference point is taken as potential energy equal to zero.

In the final state, you have kinetic energy and potential since the car has climbed 1.5[m] of the hill. Elastic energy is not available since there are no springs.

E₂ = mechanical energy at final state [J]

E_{2}=E_{kin}+E_{pot}

Now we can use the first statement to get the first equation:

E_{1}+W_{1-2}=E_{2}

where:

W₁₋₂ = work from the state 1 to 2.

E_{k}=\frac{1}{2} *m*v^{2} \\

E_{pot}=m*g*h

where:

h = elevation = 1.5 [m]

g = gravity acceleration = 9.81 [m/s²]

70 - 12 = \frac{1}{2}*2*v^{2}+2*9.81*1.5

58 = v^{2} +29.43\\v^{2} =28.57\\v=\sqrt{28.57}\\v=5.34[m/s]

4 0
3 years ago
Can someone help asap plz don’t understand how to do this
shtirl [24]
Homie I don’t know either‍♀️Drop out of schl ig ;-;
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