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3 years ago

8

A student drops a ball off the top of building and records that the ball takes 3.32s to reach the ground (g = 9.8 m/s^2). What i

s the ball speed just before hitting the ground?

1 answer:

slega [8]3 years ago

7 0

Answer:

Explanation:

Here's what we know because it was given to us:

a = -9.8 m/s/s and

time = 3.32 seconds

Here's what we know because we rock physics:

v₀ = 0 (because the object was held still before it was dropped).

Here's the equation that ties all that info together in a single one-dimensional equation:

v = v₀ + at

Filling in and solving for v:

v = 0 + (-9.8)(3.32) and

v = -33m/s

The velocity is negative because the object is moving downwards and up is positive (but you knew that already too!)

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A motorcycle is following a car that is traveling at constant speed on a straight highway. Initially, the car and the motorcycle

xz_007 [3.2K]

Answer:

a) $\Delta{t} = 5.39s$

b) the motorcycle travels 155 m

Explanation:

Let $t_2-t_1 = \Delta{t}$ , then consider the equation of motion for the motorcycle (accelerated) and for the car (non accelerated):

$v_{m2}=v_0+a\Delta{t}\\x+d=(\frac{v_0+v_{m2}}{2} )\Delta{t}\\v_c = v_0 = \frac{x}{\Delta{t}}$

where:

$v_{m2}$ is the speed of the motorcycle at time 2

$v_{c}$ is the velocity of the car (constant)

$v_{0}$ is the velocity of the car and the motorcycle at time 1

d is the distance between the car and the motorcycle at time 1

x is the distance traveled by the car between time 1 and time 2

Solving the system of equations:

$\left[\begin{array}{cc}car&motorcycle\\x=v_0\Delta{t}&x+d=(\frac{v_0+v_{m2}}{2}}) \Delta{t}\end{array}\right]$

$v_0\Delta{t}=\frac{v_0+v_{m2}}{2}\Delta{t}-d \\\frac{v_0+v_{m2}}{2}\Delta{t}-v_0\Delta{t}=d\\(v_0+v_{m2})\Delta{t}-2v_0\Delta{t}=2d\\(v_0+v_0+a\Delta{t})\Delta{t}-2v_0\Delta{t}=2d\\(2v_0+a\Delta{t})\Delta{t}-2v_0\Delta{t}=2d\\a\Delta{t}^2=2d\\\Delta{t}=\sqrt{\frac{2d}{a}}=\sqrt{\frac{2*58}{4}}=\sqrt{29}=5.385s$

For the second part, we need to calculate x+d, so you can use the equation of the car to calculate x:

$x = v_0\Delta{t}= 18\sqrt{29}=96.933m\\then:\\x+d = 154.933$

3 0

3 years ago

Which scientific activities will Juno conduct on its trip to Jupiter? Check all that apply.

Arisa [49]

Explanation:

Hope this helps,

Juno entered a polar orbit of Jupiter on July 5th 2016 UTC, to begin a scientific investigation of the planet. After completing its mission, Juno will be intentionally deorbited into Jupiters atmosphere. Junos mission is to measure Jupiters composition, gravitational field, magnetic field, and polar magnetosphere.

5 0

2 years ago

Read 2 more answers

When light of wavelength 240 nm falls on a cobalt surface, electrons having a maximum kinetic energy of 0.17 eV are emitted. Fin

dusya [7]

Answer:

(a) 5.04 eV (B) 248.14 nm (c) $1.21\times 10^{15}Hz$

Explanation:

We have given Wavelength of the light \lambda = 240 nm

According to plank's rule ,energy of light

$E = h\nu = \frac{hc}{}\lambda$

$E = h\nu = \frac{6.67\times 10^{-34} J.s\times 3\times 10^{8}m/s}{ 240\times 10^{-9} m\times 1.6\times 10^{-19}J/eV}= 5.21 eV$

Maximum KE of emitted electron i= 0.17 eV

Part( A) Using Einstien's equation

$E = KE_{max}+\Phi _{0}$ , here $\Phi _0$ is work function.

$\Phi _{0}=E - KE_{max}$ = 5.21 eV-0.17 eV = 5.04 eV

Part( B) We have to find cutoff wavelength

$\Phi _{0} = \frac{hc}{\lambda_{cuttoff}}$

$\lambda_{cuttoff}= \frac{hc}{\Phi _{0} }$

$\lambda_{cuttoff}= \frac{6.67\times 10^{-34} J.s\times 3\times 10^{8}m/s}{5.04 eV\times 1.6\times 10^{-19}J/eV }=248.14 nm$

Part (C) In this part we have to find the cutoff frequency

$\nu = \frac{c}{\lambda_{cuttoff}}= \frac{3\times 10^{8}m/s}{248.14 \times 10^{-19} m }= 1.21\times 10^{15} Hz$

5 0

3 years ago

What happens when two forces act in the same direction?

Wewaii [24]

Using the picture provided the forces are added together, because they are putting force on an object the same direction, thus the forces are added.

5 0

2 years ago

Read 2 more answers

What speed would a proton need to achieve in order to circle Earth 1790 km above the magnetic equator, where the Earth's mag- ne

irinina [24]

Answer:

The velocity is 31.25 m/s and direction is toward west.

Explanation:

Given that,

Distance $h= 1790 km = 1.790\times10^{6}\ m$

Magnetic field $B=4\times10^{-8}\ T$

Mass of proton $m=1.673\times10^{-21}\ Kg$

Radius of earth $R =6.38\times10^{6}\ m$

Radius of orbit $r=R+h$

$r=6.38\times10^{6}+1.790\times10^{6}$

$r=8170000\ m$

We need to calculate the speed

Using formula of magnetic field

$Bvq=\dfrac{mv^2}{r}$

$v=\dfrac{Bqr}{m}$

Put the value into the formula

$v=\dfrac{4\times10^{-8}\times1.6\times10^{-19}\times8170000}{1.673\times10^{-21}}$

$v=31.25\ m/s$

Hence, The velocity is 31.25 m/s and direction is toward west.

6 0

3 years ago