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Rufina [12.5K]
3 years ago
7

If a car takes a banked curve at less than the ideal speed, friction is needed to keep it from sliding toward the inside of the

curve (a real problem on icy mountain roads). (a) Calculate the ideal speed in (m/s) to take a 110 m radius curve banked at 15°. 16.99 Correct: Your answer is correct. m/s
Physics
1 answer:
exis [7]3 years ago
3 0

Answer:

ideal speed(v)≅17m/s

Explanation:

Given: Radius, r = 110 m

Angle(α) = 15°

g=9.81m/s^{2}

Find; (a) The ideal speed, v.

tanα=\frac{v^{2} }{rg}

make V subject of the formula;

v=\sqrt{rgtan\alpha }

v=\sqrt{110 * 9.81* tan15}

v=\sqrt{289.144}

v≅17m/s

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You and your friends are doing physics experiments on a frozen pond that serves as a frictionless, horizontal surface. Sam, with
Misha Larkins [42]

Answer:

Sam's and Abigail speeds before colliding were a. 12.34 m/s and b. 2.86 m/s, respectively. Their total kinetic energy was diminished by c. 1484.42 J, approximately

Explanation:

By conservation of momentum, we have

\Delta \vec{p} = 0\\\\m\vec{v}_i = m\vec{v}_f\\m_S v_S\hat{i} + m_A v_A\hat{j} = m_S \times 7 \cos{(40)} \hat{i} + m_S \times 7 \sin{(40)} \hat{j} + m_A \times 9.4 \cos{(18)} \hat{i} - m_A \times 9.4 \sin{(18)} \hat{j}.

Writing for each direction at a time,

m_S v_S = 7m_S \cos{(40)} +9.4 m_A \cos{(18)}\\v_S = 7 \cos{(40)} + 9.4 \frac{m_A}{m_S} \cos{(18)} = 7 \cos{(40)} + 9.4\times\frac{57}{73} \cos{(18)} \approx \mathbf{12.34}\\m_A v_A = 7m_S\sin{(40)} - 9.4m_A\sin{(18)}\\v_A = 7\frac{m_S}{m_A}\sin{(40)} - 9.4\sin{(18)} = 7\times\frac{73}{57}\sin{(40)} - 9.4\sin{(18)} \approx \mathbf{2.86}.

Their kinetic energy changed by

K_f - K_i = \left( \frac{1}{2}m_s v_{fs}^2 + \frac{1}{2}m_a v_{fa}^2 \right) - \left( \frac{1}{2}m_s v_{is}^2 + \frac{1}{2}m_a v_{ia}^2 \right) =  \left( \frac{1}{2}\times 73 \times 7^2 + \frac{1}{2}\times 57 \times 9.4^2 \right) - \left( \frac{1}{2}\times 73 \times 12.34^2  + \frac{1}{2}\times 57 \times 2.86^2 \right) \approx \mathbf{-1484.418 J}.

3 0
3 years ago
A sled starts from rest at the top of a hill and slides down with a constant acceleration. At some later time it is 14.4 m from
Alenkasestr [34]

Answer:

V₁  = 5.6 m/s

V₂ = 7.2 m/s

V₃ = 8.8 m/s

Explanation:

Average velocity: Average velocity can be defined as the ratio of the total  displacement to the total time taken. The S.I unit of Average velocity is m/s.

For the first 2 s,

V₁ = Δd₁/t

Where V₁  = Average velocity for the first 2 s

Where Δd₁= distance, t = time

Δd₁ = 25.6-14.4 = 11.2 m t = 2 s

V₁ = 11.2/2

V₁ = 5.6 m/s

For the second 2 s,

V₂ =Δd₂/t

Where V₂ = average velocity for the second 2 s.

Δd₂= 40-25.6 = 14.4 m, t= 2 s

V₂ = 14.4/2

V₂ = 7.2 m/s

For the last 2 seconds,

V₃ =Δd₃/t

Where V₃ = average velocity for the last 2 s

where Δd₃ = 57.6- 40 = 17.6 m, t = 2 s

V₃ = 17.6/2

V₃ = 8.8 m/s.

8 0
4 years ago
Describe 2 ways that mass and weight are the same
navik [9.2K]
Anything that has mass has weight and anything that has weight has mass simple.
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4 years ago
What is the on ohooke benden
-Dominant- [34]

Answer:Work is the energy required to move an object from one point to another. while power is the energy transferred per unit time.

6 0
3 years ago
An object experiences an acceleration of 6.8m/s2.As a result,it accelerates from rest to 24m/s.How much distance did it travel d
Hitman42 [59]

The distance covered by the object is 42.4 m

Explanation:

The motion of the object is a uniformly accelerated motion (at constant acceleration), therefore we can use the following suvat equation:

v^2 -u^2 = 2as

where

v is the final velocity

u is the initial velocity

a is the acceleration

s is the distance covered

For the object in this problem, we have:

u = 0 (it starts from rest)

v = 24 m/s (final velocity)

a=6.8 m/s^2

Solving for s, we find the distance travelled by the object:

s=\frac{v^2-u^2}{2a}=\frac{24^2-0}{2(6.8)}=42.4 m

Learn more about accelerated motion:

brainly.com/question/9527152

brainly.com/question/11181826

brainly.com/question/2506873

brainly.com/question/2562700

#LearnwithBrainly

8 0
3 years ago
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