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Vanyuwa [196]
2 years ago
15

1.) Starting from rest, a car accelerates at 6.52 m/s^2 for 3.80 s. Determine the distance traveled by the car during this time.

Physics
1 answer:
inysia [295]2 years ago
3 0

Answer:

47.1 m

Explanation:

First, write the information you're given.

v₀ = 0 m/s

a = 6.52 m/s²

t = 3.80 s

Identify the variable you're looking for:

Find: Δx

Next, find the kinematic equation you need to solve.  One trick is to identify which variable you <em>don't</em> have, then find the equation that doesn't use that variable.  In this case, we don't have the final velocity v, so the equation we should use is:

Δx = v₀ t + ½ at²

Plug in the values and solve:

Δx = (0 m/s) (3.80 s) + ½ (6.52 m/s²) (3.80 s)²

Δx = 47.1 m

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A rock is placed into a graduated cylinder containing 80 mL of water. What is the volume of the rock if the water level rises to
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2 years ago
A centrifuge used in DNA extraction spins at a maximum rate of 7000rpm producing a "g-force" on the sample that is 6000 times th
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Answer:

A) a = 73.304 rad/s²

B) Δθ = 3665.2 rad

Explanation:

A) From Newton's first equation of motion, we can say that;

a = (ω - ω_o)/t. We are given that the centrifuge spins at a maximum rate of 7000rpm.

Let's convert to rad/s = 7000 × 2π/60 = 733.04 rad/s

Thus change in angular velocity = (ω - ω_o) = 733.04 - 0 = 733.04 rad/s

We are given; t = 10 s

Thus;

a = 733.04/10

a = 73.304 rad/s²

B) From Newton's third equation of motion, we can say that;

ω² = ω_o² + 2aΔθ

Where Δθ is angular displacement

Making Δθ the subject;

Δθ = (ω² - ω_o²)/2a

At this point, ω = 0 rad/s while ω_o = 733.04 rad/s

Thus;

Δθ = (0² - 733.04²)/(2 × 73.304)

Δθ = -537347.6416/146.608

Δθ = - 3665.2 rad

We will take the absolute value.

Thus, Δθ = 3665.2 rad

8 0
3 years ago
the frequency of a beam of uv light is 1.0 ×10 ^15hz what is the energy in one quantum of this light express it in ev? ​
kap26 [50]

Answer:

4.14 eV

Explanation:

f = 1.0 ×10^15 Hz

h= 6.63×10^-34 J s (  this is called PLANCK 'S CONSTANT)

ENEGY = E = ?

E = hf  ( THIS IS FORMULA FOR ENERGY OF ONE QUANTA OR ONE PHOTON )

E= 6.63×10^-34×1.0 ×10^15

E = 6.63×10^-19 J

As 1eV = 1.6×10^-19 J so changing energy in eV from joules we will divide energy by 1.6×10^-19

hence E in eV = 6.63×10^-19/(1.6×10^-19)

          E = 4.14 eV

7 0
3 years ago
A 1600 kg sedan goes through a wide intersection traveling from north to south when it is hit by a 2000 kg SUV traveling from ea
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Answer:

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Explanation:

4 0
2 years ago
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