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Lady_Fox [76]
3 years ago
14

"The distance between the front and rear wheels of a 2280 kg car is 3.15 m. The car’s center of gravity lies in the middle of th

e car, 1.05 m behind the front wheels. When the car is at rest, what is the ratio of the force exerted by the ground on the front wheels to that exerted on the rear wheels?"

Physics
1 answer:
weeeeeb [17]3 years ago
3 0

Answer:

2

Explanation:

Hello to solve this problem we must initially draw the free body diagram of the car, and remember that for a body to be in balance the sum of its moments must be zero.

When this equation is raised, we can find the relationship between the forces of the front and rear wheel, and the complete procedure is attached.

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A baseball is thrown straight up from a building that is 25 meters tall with an initial velocity v = 10 m/s. How fast is it goin
Yanka [14]

Answer:-24,5m/s

Explanation: what we have here is a UALM with these gravity as acceleration (-9.8 m/s^2). The initial position is 25 m and initial speed is 10m/s.

Speed and gravity are increasing in the opposite direction, speed upwards and gravity downwards, while the position is also upwards, depending on your reference system.

The first thing I need to know is the maximum high it will reach.

Hmax=- S(0)^2/2g=

S= speed.

0= initial

G= gravity

Hm= 100/19,6= 5.1 m

So, the ball will go 5,1 m higher than the initial position, and from there it will fall free.

Then, I need to know how long it takes to fall. For that we use UALM equation:

X(t)= X(0) + S(0)*t + (A*t^2)/2.

X: position

S: speed

A: acceleration

T:time

0: initial

0 = 25m +10*t -(9.8 * t^2)/2

Solving the quadratic equation we get

T= 3,5 sec. ( Negative value for time is impossible)

So now we know that the ball to go up and then fall needs 3,5 sec.

Let's see how long it takes to go up:

30,1=25+10*t-4,9*t^2

0=-5,1+10*t-4,9*t^2

T= 1 sec. So it will take 1 sec to the ball to reach the maximum high and 0=speed and then it'll fall during the resting 2,5 sec

Finally, to know the speed just before it touches the ground, we use the following formula:

A= (St-S0)/t

-9.8m/s^2 = (St- 0m/s)/ 2,5s

-24,5 m/s= St

-24,5 m/s is the speed at 3,5 sec, which is the time just before falling

3 0
3 years ago
Which of these would make the best telescope?
Simora [160]

Answer:A i think or D but its not c or b

Explanation:

5 0
2 years ago
The work that a force does by acting on an object is equal to what?
maw [93]
2nd and only 2nd option is right
3 0
3 years ago
Help please!!!
exis [7]

Answer:

5 years worth of work (aka all of the homework i currently have)

3 0
3 years ago
The temperature of a 700.96 gram piece of metal falls 120⁰C and in the process releases 2001 Joules of energy. What is the speci
olga nikolaevna [1]

Answer:

The specific heat for the metal is 0.466 J/g°C.

Explanation:

Given,

Q = 1120 Joules

mass = 12 grams

T₁ = 100°C

T₂ = 300°C

The specific heat for the metal can be calculated by using the formula

Q = (mass) (ΔT) (Cp)

ΔT = T₂ - T₁ = 300°C  - 100°C   = 200°C

Substituting values,

1120 = (12)(200)(Cp)

Cp = 0.466 J/g°C.

Therefore, specific heat of the metal is 0.466 J/g°C.

8 0
3 years ago
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