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Whitepunk [10]
4 years ago
12

The four wheels of a car are connected to the car's body by spring assemblies that let the wheels move up and down over bumps an

d dips in the road. When a 68 kg (about 150 lb) person sits on the left front fender of a small car, this corner of the car dips by about 1.2 cm (about 12 in). If we treat the spring assembly as a single spring, what is the approximate spring constant?
Physics
1 answer:
Marrrta [24]4 years ago
7 0

Answer:k=55.590 KN/m

Explanation:

Given

mass of person\left ( m\right )=68 kg

car dips about 1.2 cm

We know

F=kx

Where k=combined  spring constant

mg=kx

k=\frac{mg}{x}

k=\frac{68\times 9.81}{1.2\time 10^{-2}}

k=55.590 KN/m

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4 years ago
a 550 kg car accelerates from 15 m/s to 25 m/s in 15 s by applying a constant force. how large of a force is exerted? Please sho
Novay_Z [31]
You'll be using the equation f = m a, or force = mass x acceleration

First, you have to find the acceleration. The acceleration needed is the average acceleration over the 15 seconds is accelerated. So, you take the change in speed (25m/s - 15m/s) to get a change of 10m/s. 
The average acceleration (acceleration per second) is found by dividing total acceleration by the time it took. So, it's 10 / 15, which equals .6. This is a, your acceleration
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F = 550 x .6
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3 0
3 years ago
Which of these would NOT transfer and change electrical energy into another type of energy in a closed circuit? A) fan B) air C)
Nitella [24]
The answer is B) air.
7 0
4 years ago
Read 2 more answers
A kangaroo jumps up with an initial velocity of 36 feet persecond from the ground (assume its starting height is 0 feet).Use the
kkurt [141]

Given

Initial velocity:

36 ft/s

Initial height:

0 ft

Vertical motion model:

h(t) = -16t^2 + ut + s

v = initial velocity

s = is the height

Procedure

We are going to use the model provided for the vertical motion.

\begin{gathered} h(t)=-16t^2+36t+0 \\ h(t)=-16t^2+36t \end{gathered}

We know that at the maximum height the final velocity is 0.

Then we will use the following expression to calculate the maximum height:

\begin{gathered} v^2_f=v^2_o-2ah_{\max } \\ 0=v^2_o-2ah_{\max } \\ 2ah_{\max }=v^2_o \\ h_{\max }=\frac{v^2_o}{2a} \\ h_{\max }=\frac{(36ft/s)^2}{2\cdot32ft/s^2} \\ h_{\max }=20.25\text{ ft} \end{gathered}

Now for time:

\begin{gathered} 20.25=-16t^2+36t \\ 16t^2-36t+20.25=0 \end{gathered}

Solving for t,

\begin{gathered} t_1=2.25 \\ t_2=0 \end{gathered}

The total time the kangaroo takes in the air is 2.3s.

3 0
1 year ago
A 0.8 kg object displaces 500 mL of water. What is its specific gravity?
Airida [17]
The specific gravity is how the density of the object compares to the density of water. Water's density is 1gram per milliliter. We just need to figure out the density of the object.

The object is .8 kg and it displaces 500mL of water, so the density is the mass divided by the volume. Since the density of water is given in grams, we have to convert the objects mass from kg to g and then we can get the density.

.8kg * 1000g/kg = 800 grams

So

800g/500ml = 1.6grams/mL this is the density.

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8 0
4 years ago
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