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3 years ago

A race car's velocity increases from +28 m/s to +36 m/s over a 2.0-s time interval. What is the car's

Physics

1 answer:

BigorU [14]3 years ago

3 0

Answer:

$4 m/s^{2}$

Explanation:

Acceleration, $a=\frac {v-u}{t}$

Where v and u are the final and initial velocities of the race car respectively, t is the time taken for the race car to attain velocity of 36 m/s.

Substituting 36 m/s for v, 28 m/s for u and 2 s for t then

$a=\frac {36 m/s-28 m/s}{2}=4 m/s^{2}$

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Consider a hydrogen atom in the n = 1 state. The atom is placed in a uniform B field of magnitude 2.5 T. Calculate the energy di

dlinn [17]

Answer:

$E=29\times 10^{-5}eV$

Explanation:

For n-=1 state hydrogen energy level is split into three componets in the presence of external magnetic field. The energies are,

$E^{+}=E+\mu B$ ,

$E^{-}=E-\mu B$ ,

$E^{0}=E$

Here, E is the energy in the absence of electric field.

And

$E^{+} and E^{-}$ are the highest and the lowest energies.

The difference of these energies

$\Delta E=2\mu B$

$\mu=9.3\times 10^{-24}J/T$ is known as Bohr's magneton.

B=2.5 T,

Therefore,

$\Delta E=2(9.3\times 10^{-24}J/T)\times 2.5 T\\\Delta E=46.5\times 10^{-24}J$

Now,

$Delta E=46.5\times 10^{-24}J(\frac{1eV}{1.6\times 10^{-9}J } )\\Delta E=29.05\times 10^{-5}eV\\Delta E\simeq29\times 10^{-5}eV$

Therefore, the energy difference between highest and lowest energy levels in presence of magnetic field is $E=29\times 10^{-5}eV$

6 0

3 years ago

Michael Jordan, el célebre basquetbolista, ganó el torneo de clavadas de la NBA en 1988. Para lograr la hazaña saltó 1.35 metros

kozerog [31]

(a) 0.40 s

First of all, let's find the initial speed at which Jordan jumps from the ground.

The maximum height is h = 1.35 m. We can use the following equation:

$v^2-u^2=2gh$

where

v = 0 is the velocity at the maximum height

u is the initial velocity

$g=-9.8 m/s^2$ is the acceleration of gravity

Solving for u,

$u=\sqrt{-2gh}=\sqrt{-2(-9.8)(1.35)}=5.14 m/s$

The time needed to reach the maximum height can now be found by using the equation

$v=u+gt$

Solving for t,

$t=\frac{v-u}{g}=\frac{0-5.14}{-9.8}=0.52s$

Now we can find the velocity at which Jordan reaches a point 20 cm below the maximum height, so at a height of

h' = 1.35 - 0.20 = 1.15 m

Using again the equation

$v'^2-u^2=2gh'$

we find

$v'=\sqrt{u^2+2gh}=\sqrt{5.14^2+2(-9.8)(1.15)}=1.97 m/s$

And the corresponding time is

$t'=\frac{v'-u}{g}=\frac{1.97-5.14}{-9.8}=0.32s$

So the time to go from h' to h is

$\Delta t = t-t'=0.52-0.32=0.20 s$

And since we have also to take into account the fall down (after Jordan reached the maximum height), which is symmetrical, we have to multiply this time by 2 to get the total time of permanence in the highest 20 cm of motion:

$\Delta t=2\cdot 0.20 = 0.40 s$

(b) 0.08 s

This part is easier since we need to calculate only the velocity at a height of h' = 0.20 m:

$v'^2-u^2=2gh'$

$v'=\sqrt{u^2+2gh}=\sqrt{5.14^2+2(-9.8)(0.20)}=4.74 m/s$

And the corresponding time is

$t'=\frac{v'-u}{g}=\frac{4.74-5.14}{-9.8}=0.04s$

So this is the time needed to go from h=0 to h=20 cm; again, we have to take into account the motion downwards, so we have to multiply this by 2:

$\Delta t = 2\cdot 0.04 =0.08 s$

8 0

3 years ago

A 0.12 kg body undergoes simple harmonic motion of amplitude 8.5 cm and period 0.20 s. (a) What is themagnitude of the maximum f

Neporo4naja [7]

Answer:

a)F=698.83 N

b)K=8221.56 N/m

Explanation:

Given that

mass ,m = 0.12 kg

Amplitude ,A= 8.5 cm

time period ,T = 0.2 s

We know that

$T=\dfrac{2\pi}{\omega}$

${\omega}=\dfrac{2\pi }{0.2}\ rad/s$

${\omega}=31.41\ rad/s$

We know that

${\omega}^2=m\ K$

K=Spring constant

$K=\dfrac{\omega^2}{m}$

$K=\dfrac{31.41^2}{0.12}\ N/m$

K=8221.56 N/m

The maximum force F

F= K A

F= 8221.56 x 0.085 N

F=698.83 N

a)F=698.83 N

b)K=8221.56 N/m

3 0

3 years ago

This problem has been solved!See the answerCommunication with submerged submarines via radio waves is difficult because seawater

mash [69]

Answer:

4000 km

Explanation:

as we know velocity of electromagnetic wave is c

$c = 3 * 10^8 m/s$

frequency given (f) = 76 Hz

wavelength ?

using

$c =f$ λ

λ = $\frac{c}{f}$

λ = $\frac{3 * 10^8}{76}$ ≈4000 km

so final answer λ = 4000km

8 0

3 years ago

Air (14.5 lb) undergoes a polytropic process in a closed system from p1 = 80 lbf/in2, υ1 = 4 ft3/lb to a final state where p2 =

Yanka [14]

The energy transfer in terms of work has the equation:

W = mΔ(PV)

To be consistent with units, let's convert them first as follows:

P₁ = 80 lbf/in² * (1 ft/12 in)² = 5/9 lbf/ft²
P₂ = 20 lbf/in² * (1 ft/12 in)² = 5/36 lbf/ft²
V₁ = 4 ft³/lbm
V₂ = 11 ft³/lbm

W = m(P₂V₂ - P₁V₁)
W = (14.5 lbm)[(5/36 lbf/ft²)(4 ft³/lbm) - (5/9 lbf/ft²)(11 lbm/ft³)]
W = -80.556 ft·lbf

In 1 Btu, there is 779 ft·lbf. Thus, work in Btu is:
W = -80.556 ft·lbf(1 Btu/779 ft·lbf)
<em>W = -0.1034 BTU</em>

4 0

3 years ago