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HACTEHA [7]
3 years ago
11

A mechanic test driving a car that she has just given a tune-up accelerates from rest to 50.0 m/s in 9.8 s. How far (in meters)

does she travel in that time?
Physics
1 answer:
Stels [109]3 years ago
6 0

Answer:

245 m

Explanation:

v = at + v₀

50.0 m/s = a (9.8 s) + 0 m/s

a = 5.10 m/s²

x = x₀ + v₀ t + ½ at²

x = 0 m + (0 m/s) (9.8 s) + ½ (5.10 m/s²) (9.8 s)²

x = 245 m

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That's the (northern hemisphere) summer solstice, namely the one in June.
4 0
3 years ago
What minimum speed must the block have at the base of the 70 m hill to pass over the pit at the far (right-hand) side of that hi
Drupady [299]

Answer:

initial velocity is v = 4.95 m / s

Explanation:

To solve this exercise we use the projectile launch ratios, when the block leaves the hill its speed is horizontal, let's find the time it takes to fall to the other point.

Initial vertical velocity is zero

          y = y₀ + v_{oy} t - ½ g t²

          y-y₀ = 0 -1/2 g t²

          t = \sqrt{ \frac{ 2(y_o -y)}{g} }

calculate

          t = \sqrt{ \frac{2 ( 70-50)}{9.8} }

          t = 2.02 s

with this time we can substitute in the horizontal displacement equation

          x = v₀ₓ t

          v₀ₓ = x / t

suppose that the distance between the two points is x = 10 m

          v₀ₓ = 10 / 2.02

          v₀ₓ = 4.95 m / s

initial velocity is v = 4.95 m / s

4 0
3 years ago
How many protons are in the radioactive isotope 40/19K?
trasher [3.6K]
First we have to establish that the number of protons is equivalent to the atomic number of element. Here I am assuming that you are referring to Potassium (K) - 40. Potassium, stable or unstable has 19 protons.
8 0
3 years ago
A car drives around a curve with radius 539 m at a speed of 32.0 m/s. The road is banked at 5.00°. The mass of the car is 1.40 ×
HACTEHA [7]

Answer:

f_r = 150.47 N

Explanation:

given,

r = 539 m

v = 32 m/s

road banked at = 5°

∑ F_x

\dfrac{mv^2}{r}= N sin \theta + f_r cos \theta

∑ F_y = 0

0 = N cos \theta - f_r sin \theta - mg

N = \dfrac{f_rsin \theta + mg}{cos \theta}

\dfrac{mv^2}{r}= (\dfrac{f_rsin \theta + mg}{cos \theta})sin \theta + f_r cos \theta

              = f_r sin \theta tan \theta + mg tan \theta + f_r cos \theta

        f_r = \dfrac{\dfrac{mv^2}{r}- mg tan\theta}{sin\theta tan \theta + cos \theta}

         f_r = \dfrac{\dfrac{1.4\times 10^3 \times 32^2}{539}- 1.4\times 10^{3}\times 9.8 \times 0.087}{0.087 \times 0.087 + 0.996}

f_r = 150.47 N

8 0
3 years ago
Class II levers like ankles and wheelbarrows are useful because they provide mechanical advantage, by amplifying the input force
marusya05 [52]

Answer:

The solution and the explanation are in the Explanation section.

Explanation:

According to the diagram that is in the attached image, the EFFORT force at point A and the load is at O point. The torque due to weight is:

TA = W * (a * cosθ)

The torque due to effort at C point is:

TC = E * (b * cosθ)

The net torque is equal to 0, we have:

Tnet = 0

W * (a * cosθ) - E * (b * cosθ) = 0

E=W\frac{a}{b}

From the figure, you can observe that a/b < 1, thus E < W

8 0
3 years ago
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