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mash [69]
3 years ago
14

KE 1/2mv^2 is a for kinetic energy (KE) in terms of mass, m, and the velocity, v, of a moving object. Which choice shows a corre

ct way to express the mass in terms of the kinetic energy and velocity?
A. m=2KE/v^2, where v≠0
B. m=KE/2v^2, where v≠0
C. m √2KE/v, where v≠0
D. m= √KE/2v, where v≠0
Physics
2 answers:
aivan3 [116]3 years ago
7 0

Answer:Kinetic energy is directly proportional to the mass of the object and to the square of its velocity: K.E. = 1/2 m v2. If the mass has units of kilograms and the velocity of meters per second, the kinetic energy has units of kilograms-meters squared per second squared.

motikmotik3 years ago
6 0

energy = (1/2)*m*(v^2)

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During the experiment if you could double the breakaway magnetic force with all other quantities left unchanged, what is the new
sergiy2304 [10]
There are some missing information in the question.
However, since you are talking about magnetic force, I think you refer to the Lorentz force. When a particle of charge q and velocity v is immersed in a magnetic field of intensity B, the force acting on the particle is:
F=qvBsin\theta
where \theta is the angle between the magnetic field and the direction of the particle.
Therefore, if force F is doubled, then also the velocity v must be double of its initial value:
v=2v_0
6 0
3 years ago
A television camera lens has a 17-cm focal length and a lens diameter of 6.0 cm. what is its number?
IRINA_888 [86]

Answer:

= 2.83

Explanation:

F number (N) is given by the formula;

  F- number = f/D

where f = focal length of lens and D = diameter of the aperture  

Therefore;

F number = 17 cm/6 cm

                <u> = 2.83</u>

3 0
3 years ago
Assume it takes 10 J to stretch a spring 10 cm beyond its natural length. Find the work required (in Joules) to stretch the spri
Lady bird [3.3K]

Answer:

W = 30 J

Explanation:

given,

Work done = 10 J

Stretch of spring, x = 0.1 m

We know,

dW = F .dx

we know, F = k x

\int dW = \int_0^{0.1} k.x dx

W = \int_0^{0.1} k.x dx

W = k[\dfrac{x^2}{2}]_0^{0.1}

10 = k\dfrac{0.1^2}{2}

k = 2000

now, calculating Work done by the spring when it stretched to 0.2 m from 0.1 m.

W = \int_{0.1}^{0.2} 2000 x dx

W = 2000 [\dfrac{x^2}{2}]_{0.1}^{0.2} dx

W = 1000 x 0.03

W = 30 J

Hence, work done is equal to 30 J.

4 0
3 years ago
Canadian Olivia Oliver broke the Guinness World Record for the fastest spin on ice skates on January 2015 with an angular veloci
Shalnov [3]

Answer:

<em>a) 2.63 : 1</em>

<em>b) 143.13 rpm</em>

Explanation:

initial angular velocity ω' = 130 rpm

final angular velocity ω = 342 rpm

recall that angular momentum = ωI

where I is the moment of inertia.

for the initial spinning condition, we take moment of inertial = I'

for final spinning condition, we take moment of inertia = I

initial angular momentum = ω'I' = 130 I'

final angular momentum = ωI = 342 I

according to conservation of angular momentum, initial angular momentum must be equal to the final angular momentum, therefore

342 I = 130 I'

ratio of initial moment of inertia to final moment of inertia = I'/I

==> I'/I = 342/130 ≅ <em>2.63 : 1</em>

b) to achieve a final angular velocity of of 375 rpm, her initial velocity will have to be

2.63 = 375/ω'

ω' = 375/2.63 = <em>143.13 rpm</em>

6 0
3 years ago
A stunt driver wants to make his car jump over 8 cars parked side by side below a horizontal ramp.
Genrish500 [490]

<u>Answer:</u>

a) Minimum speed must he drive off the horizontal ramp = 39.78 m/s

b) Minimum speed must he drive off the horizontal ramp with 7° above the horizontal  = 23.93 m/s

<u>Explanation:</u>

a) The height of ramp = 1.5 meter

   Horizontal distance he must clear = 22 meter

   The car is having horizontal motion and vertical motion. In case of vertical motion the acceleration on the car is acceleration due to gravity.

   We have equation of motion, s= ut+\frac{1}{2} at^2, s is the displacement, u is the initial velocity, a is the acceleration and t is the time.

 In case of vertical motion initial velocity = 0 m/s, acceleration = 9.8 m/s^2, we need to calculate time when displacement = 1.5 meter.

 1.5=0*t+\frac{1}{2} *9.8*t^2\\ \\ t = 0.553 seconds

So the car has to cover a distance of 22 meter in 2.119 seconds.

 So minimum speed required = 22/0.553 = 39.78 m/s

 Minimum speed must he drive off the horizontal ramp = 39.78 m/s

b) When the take of angle is 7⁰ the vertical speed of car is not zero = V sin 7 = 0.122 V

 So the in case of vertical motion we have initial velocity = 0.122 V, S = -1.5 meter( below ramp), acceleration = -9.8 m/s^2

Substituting

     -1.5=0.122V*t-\frac{1}{2} *9.8*t^2\\ \\ 4.9t^2-0.122Vt-1.5=0

In case of horizontal motion

    Horizontal speed of car = V cos 7 = 0.993V

    So it has to travel 22 meter in t seconds

            0.993Vt = 22, Vt = 22.155 m

    Substituting in the equation 4.9t^2-0.122Vt-1.5=0

    We will get 4.9t^2-0.122*22.155-1.5=0\\ \\ t = 0.926 seconds

   Speed required = 22.155/0.926 = 23.93 m/s

  Minimum speed must he drive off the horizontal ramp with 7° above the horizontal  = 23.93 m/s

7 0
3 years ago
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