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castortr0y [4]
3 years ago
12

Hi, I am having some issues with this physics problem, and it is vital for me to get this problem solved. Can anyone please give

me the answer to the problem in the image and tell me how they got the answer with a step by step explanation? I would really appreciate help! ​

Physics
1 answer:
mafiozo [28]3 years ago
6 0

Answer:

U = 30 m/s

a = 3 m/s²

Explanation:

"A car accelerating uniformly from rest reaches a maximum speed of U in 10 s.  It then moves with that speed for an additional 20 s.  The distance covered by the car in the 30 s interval is 750 m.  Find U and the acceleration of the car in the first 10 s."

During the first 10 s:

v₀ = 0 m/s

v = U m/s

t = 10 s

The distance covered in this time is the average velocity times time:

Δx = ½ (v + v₀) t

Δx = ½ (U + 0) (10)

Δx = 5U

The distance covered in the next 20 seconds is speed times time:

Δx = 20U

The total distance is 750 m:

5U + 20U = 750

25U = 750

U = 30 m/s

The acceleration during the first 10 seconds is the change in speed over change in time.

a = Δv / Δt

a = (30 m/s − 0 m/s) / 10 s

a = 3 m/s²

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Aliun [14]

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hub9hybygbgybgybgygybsbgydgbydxbgbyxdgbyxdyggdxygbyxdgybzgbydbgyzsbgydgbyzdgxbybgydzs

Explanation:

4 0
3 years ago
Read 2 more answers
A 62.0-kg athlete leaps straight up into the air from a trampoline with an initial speed of 9.6 m/s. The goal of this problem is
pochemuha

Answer:

2856.96 J

0

0

\frac{1}{2}mv_i^2+mgh_i=\frac{1}{2}mv_f^2+mgh_f

6.78822 m/s

Explanation:

v_i = Initial velocity = 9.6 m/s

g = Acceleration due to gravity = 9.81 m/s²

h = Height

The athlete only interacts with the gravitational potential energy. Air resistance is neglected.

At height y = 0

Kinetic energy

K=\frac{1}{2}mv^2\\\Rightarrow K=\frac{1}{2}\times 62\times 9.6^2\\\Rightarrow K=2856.96\ J

At height y = 0 the potential energy is 0 as

P=mgy\\\Rightarrow P=mg0=0

At maximum height her velocity becomes 0 so the kinetic energy becomes zero.

As the the potential and kinetic energy are conserved

The general equation

K_i+P_i=K_f+P_f\\\Rightarrow \frac{1}{2}mv_i^2+mgh_i=\frac{1}{2}mv_f^2+mgh_f

Half of maximum height

\\\Rightarrow mgh_i+\frac{1}{2}mv_f^2=mg\frac{h_i}{2}+\frac{1}{2}mv^2\\\Rightarrow gh_i=g\frac{h_i}{2}+\frac{1}{2}v^2\\\Rightarrow g\frac{h_i}{2}=\frac{1}{2}v^2\\\Rightarrow v=\sqrt{gh}

h_i=\frac{v_i^2}{2g}

v=\sqrt{gh}\\\Rightarrow v=\sqrt{g\times \frac{v_i^2}{2g}}\\\Rightarrow v=\sqrt{\frac{v_i^2}{2}}\\\Rightarrow v=\sqrt{\frac{9.6^2}{2}}\\\Rightarrow v=6.78822\ m/s

The velocity of the athlete at half the maximum height is 6.78822 m/s

8 0
3 years ago
As you may have experienced yourself, water alone will not remove oil from a dirty dish. Why does soap work to remove oil
Alexandra [31]

Answer:

Soap breaks up the oil into smaller drops, which can mix with the water. It works because soap is made up of molecules with two very different ends (one end of molecules are hydrophilic, so they love water; the other end of molecules is hydrophobic, so they hate water).

5 0
3 years ago
Read 2 more answers
Please help I’m so confused
Ilya [14]

<em>1.wavelength</em>

<em>2.trough</em>

<em>3.amplitude</em>

<em>4.crest</em>

5 0
3 years ago
A 2.0-cm-diameter, 15-cm-long solenoid is tightly wound with one layer of wire. A 2.5 A current through the wire generates a 3.0
sweet [91]

Answer:

d = 1.047 mm

Explanation:

given,

diameter of the wire = 2.0-cm

length of solenoid = 15 cm = 0.15

Current in the wire = I = 2.5 A

magnetic field = B = 3.0 mT

Magnetic field inside the solenoid

        B = \dfrac{\mu_0 N I}{L}

        B = \dfrac{\mu_0 N I}{L}

               N x d = l

        N = \dfrac{l}{d}

        B = \dfrac{\mu_0 \dfrac{l}{d} I}{L}

        3 \times 10^{-3} = \dfrac{4\pi \times 10^{-7}\times \dfrac{0.15}{d}\times 2.5}{0.15}

        0.45 \times 10^{-3}\ d = 4\pi \times 10^{-7}\times 0.15\times 2.5

        \ d = \dfrac{4\pi \times 10^{-7}\times 0.15\times 2.5}{0.45 \times 10^{-3}}

               d = 1.047 x 10⁻³ m

               d = 1.047 mm

diameter of the wire is d = 1.047 mm

5 0
4 years ago
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