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Illusion [34]
3 years ago
6

A car starts out traveling at 35 m/s. The car hits the brakes and decelerates at a rate of 3 m/s^2 for 5 seconds. What Distance

does the car travel during this period of time?
Physics
1 answer:
Ipatiy [6.2K]3 years ago
6 0
Answer:

Time needed: 2.5 s
Distance covered: 31.3 m

Explanation:

I'll start with the distance covered while decelerating. Since you know that the initial speed of the car is 15.0 m/s, and that its final speed must by 10.0 m/s, you can use the known acceleration to determine the distance covered by

v2f=v2i−2⋅a⋅d

Isolate d on one side of the equation and solve by plugging your values

d=v2i−v2f2a

d=(15.02−10.02)m2s−22⋅2.0ms−2

d=31.3 m

To get the time needed to reach this speed, i.e. 10.0 m/s, you can use the following equation

vf=vi−a⋅t, which will get you

t=vi−vfa

t=(15.0−10.0)ms2.0ms2=2.5 s

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3 years ago
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1. A hydraulic lift is to be used to lift a truck masses 5000 kg. If the diameter of the
zheka24 [161]

Answer:

a)   F = 4.9 10⁴ N,  b)   F₁ = 122.5 N

Explanation:

To solve this problem we use that the pressure is transmitted throughout the entire fluid, being the same for the same height

1) pressure is defined by the relation

           P = F / A

to lift the weight of the truck the force of the piston must be equal to the weight of the truck

          ∑F = 0

          F-W = 0

          F = W = mg

          F = 5000 9.8

          F = 4.9 10⁴ N

the area of ​​the pisto is

          A = pi r²

          A = pi d² / 4

          A = pi 1 ^ 2/4

          A = 0.7854 m²

pressure is

          P = 4.9 104 / 0.7854

          P = 3.85 104 Pa

2) Let's find a point with the same height on the two pistons, the pressure is the same

          \frac{F_1}{A_1} = \frac{F_2}{A_2}

where subscript 1 is for the small piston and subscript 2 is for the large piston

          F₁ = \frac{A_1}{A_2} \ F_2

the force applied must be equal to the weight of the truck

          F₁ = ( \frac{d_1}{d_2} )^2\  m g

          F₁ = (0.05 / 1) ² 5000 9.8

          F₁ = 122.5 N

7 0
2 years ago
An Earth satellite is in a circular orbit at an altitude of 500 km.
Dafna1 [17]

Answer:

Speed Unchanged.

Explanation:

As work is defined as a product of force over a distance. If the distance in altitude is constant = 500km, there's 0 change in distance and force, no work would be done by the gravitational force.

Since potential energy of the satellite is unchanged, unless there's additional internal energy source, the kinetic energy would remain unchanged, so would its speed.

8 0
2 years ago
What is the purpose of using significant figures? How does it relate to accuracy, precision, resolution, and uncertainty?
Umnica [9.8K]

Answer:

#see solution for details

Explanation:

-Uncertainty refers to an estimate of the amount by which a result may differ from this value,

-Precision refers to how closely repeated measurements agree with each other.

-Accuracy refers to how closely a measured value agrees with the correct value.

-The number of significant figures is the number of digits believed to be correct by the person doing the measuring. Therefore, choosing the correct number of significant figures reduces the deviation from the point of accuracy/uncertainty or precision and thereby reducing margin of error in the ensuing calculations.

7 0
3 years ago
If an electric current of 8.50 A flows for 3.75 hours through an electrolytic cell containing copper-sulfate (CuSO4) solution, t
muminat

Answer:

75.5g

Explanation:

From the ionic equation, we can write

CU^{2+}+SO^{2-}_{4}\\

next we find the number of charge

Note Q=it

for i=8.5A, t=3.75 to secs 3.75*60*60=13500secs

hence

Q=8.5*13500\\Q=114750C

Since one faraday represent one mole of electron which equal 96500C

Hence the number of mole produced by 114750C is

114750/96500=1.2mol

The mass of copper produced is

mol=\frac{mass}{molar mass} \\mass=mole*molar mass\\mass=1.2*63.5\\mass=75.5g

Hence the amount of copper produced is 75.5g

7 0
3 years ago
Read 2 more answers
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