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ANEK [815]
4 years ago
11

a bicyclist is moving at a speed of 5.5 m/s. she squeezes the brakes, giving her an acceleration of -2.1 m/s². how fast is the b

icyclist going after 0.3 seconds of applying the brakes?
Physics
1 answer:
Blizzard [7]4 years ago
4 0
Formula relating time, velocity and acceleration:

Vf = Vi + at (Vf = Final velocity, Vi = initial velocity)

What you're given:
Initial speed, Vi = 5.5 m/s
Acceleration, a = -2.1 m/s^2
Time, t = 0.3 s

Required:
Vf = ?

Plugging in the given information:

Vf = (5.5) + (-2.1)(0.3)
Vf = 4.87 m/s

The bicyclist's speed is 4.87 m/s after 0.3 s.
You might be interested in
A track and field athlete is running forward with a javelin at a
rosijanka [135]

Explanation:

The relative velocity is the velocity of the athlete relative to the ground plus the velocity of the javelin relative to the athlete.

v = 4.2 m/s + 10.3 m/s

v = 14.5 m/s

5 0
3 years ago
How are objects in space able to “fall” into orbit?
Zolol [24]

Answer:

MRCORRECT has answered the question

Explanation:

Newton realized that the reason the planets orbit the Sun is related to why objects fall to Earth when we drop them. The Sun's gravity pullson the planets, just as Earth's gravity pulls down anything that is not held up by some other force and keeps you and me on the ground.

7 0
3 years ago
You have been hired as a technical consultant for an early-morning cartoon series for children to make sure that the science is
katen-ka-za [31]

The initial potential energy of the wagon containing gold boxes will enable

it roll down the hill when cut loose.

The Lone Ranger and Tonto have approximately <u>5.1 seconds</u>.

Reasons:

Mass wagon and gold = 166 kg

Location of the wagon = 77 meters up the hill

Slope of the hill = 8°

Location of the rangers = 41 meters from the canyon

Mass of Lone Ranger, m₁ = 65 kg

Mass of Tonto m₂ = 66 kg

Solution;

Height of the wagon above the level ground, h = 77 m × sin(8°) ≈ 10.72 m

Potential energy = m·g·h

Where;

g = Acceleration due to gravity ≈ 9.81 m/s²

Potential energy of wagon, P.E. ≈ 166 × 9.81 × 10.72 = 17457.0912

Potential energy of wagon, P.E. ≈ 17457.0912 J

By energy conservation, P.E. = K.E.

K.E. = \mathbf{\dfrac{1}{2} \cdot m \cdot v^2}

Where;

v = The velocity of the wagon a the bottom of the cliff

Therefore;

\dfrac{1}{2} \times 166 \times v^2 = 17457.0912

v = \sqrt{\dfrac{17457.0912}{\dfrac{1}{2} \times 166} } \approx 14.5

Velocity of the wagon, v ≈ 14.5 m/s

Momentum = Mass, m × Velocity, v

Initial momentum of wagon = m·v

Final momentum of wagon and ranger = (m + m₁ + m₂)·v'

By conservation of momentum, we have;

m·v = (m + m₁ + m₂)·v'

\therefore v' = \mathbf{ \dfrac{m \cdot v}{(m + m_1 + m_2)  }}

Which gives;

\therefore v' = \dfrac{166 \times 14.5}{(166 + 65 + 66)  } \approx 8.1

The velocity of the wagon after the Ranger and Tonto drop in, v' ≈ 8.1 m/s

Time = \dfrac{Distance}{Velocity}

\mathrm{The \ time \ the\ Lone \  Ranger \  and  \ Tonto \  have,  \ t} = \dfrac{41 \, m}{8.1 \, m/s} \approx 5.1 \, s

The Lone Range and Tonto have approximately <u>5.1 seconds</u> to grab the

gold and jump out of the wagon before the wagon heads over the cliff.

Learn more here:

brainly.com/question/11888124

brainly.com/question/16492221

5 0
3 years ago
Which types of numbers does scientific notation best describe?
Travka [436]
The correct answer is
<span>c) very small and very large

Let's see this with a few examples:
1) if we have a very small number, such as
</span>0.0000000001
<span>we see that we can write it easily by using the scientific notation:
</span>1\cdot 10^{-10}
<span>2) Similarly, if we have a very large number:
</span>10000000000
<span>we see that we can write it easily by using again the scientific notation:
</span>1 \cdot 10^{10}<span>
</span>
4 0
3 years ago
A conductor carrying a current I = 16.5 A is directed along the positive x axis and perpendicular to a uniform magnetic field. A
Jet001 [13]

To solve this problem we will apply the concepts related to the Magnetic Force, this is given by the product between the current, the body length, the magnetic field and the angle between the force and the magnetic field, mathematically that is,

F = ILBsin \theta

Here,

I = Current

L = Length

B = Magnetic Field

\theta = Angle between Force and Magnetic Field

But \theta = 90\°

F = ILB

Rearranging to find the Magnetic Field,

B = \frac{F}{IL}

Here the force per unit length,

B = \frac{1}{I}\frac{F}{L}

Replacing with our values,

B = \frac{0.130N/m}{16.5}

B = 0.0078T

Therefore the magnitude of the magnetic field in the region through which the current passes is 0.0078T

6 0
3 years ago
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