A major advantage of case studies is _______

What is the acceleration of a 30.0-kg go kart if the thrust of its engine is 300.0 N?

VikaD [51]

Answer : $a=10\ m/s^2$

Explanation :

It is given that,

Mass of the engine, m = 30 kg

Thrust is equivalent to the force acting perpendicularly and it is F = 300 N

According to Newton's second law of motion :

$F = m\times a$

a is the acceleration of the engine.

$a=\dfrac{F}{m}$

$a=\dfrac{300\ N}{30\ Kg}$

$a=10\ m/s^2$

So, the acceleration of the engine is $10\ m/s^2$ .

Hence, this is the required solution.

5 0

3 years ago

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Light of a given wavelength is used to illuminate the surfacce of a metal, however, no photoelectrons are emitted. In order to c

Jlenok [28]

Answer:

B. use light of a shorter wavelength.

Explanation:

We know that

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

h= plank's constant

c= speed of light

λ= wavelength of the incident light

so, in order to have sufficient energy for for the emission of electron, the incident's radiation must have λ small enough.

B. use light of a shorter wavelength.

6 0

3 years ago

A 20,000 kg truck is traveling at 10 m/s. Calculate the truck's momentum. (p = m v)

kow [346]

Answer:

200,000 and 20,000,000

Explanation:

Substituting the values into the equation of momentum, we get that the momentum is p = mv = 20,000*10 = 200,000. Using the equation provided to solve for the force the truck experienced, we find: 20,000*10/0.01 = 20,000,000.

I hope this helps!

8 0

3 years ago

A 225 kg block is pulled by two horizontal forces. The first force is 178 N at a 41.7-degree angle and the second is 259 N at a

yawa3891 [41]

Answer:

52.9 N, 364.7 N

Explanation:

First of all, we need to resolve both forces along the x- and y- direction. We have:

- Force A (178 N)

$A_x = (178 N)(cos 41.7^{\circ})=132.9 N\\A_y = (178 N)(sin 41.7^{\circ})=118.4 N$

- Force B (259 N)

$B_x = (259 N)(cos 108^{\circ})=-80.0 N\\B_y = (259 N)(sin 108^{\circ})=246.3 N$

So the x- and y- component of the total force acting on the block are:

$R_x = A_x + B_x = 132.9 N - 80.0 N =52.9 N\\R_y = A_y + B_y = 118.4 N +246.3 N = 364.7 N$

7 0

3 years ago

The magnetic field over a certain range is given by B~ = Bx ˆı + By ˆ, where Bx = 2 T and By = 4 T. An electron moves into the

krek1111 [17]

Answer:

Explanation:

The force exerted in a magnetic field is given as

F = q (v × B)

Where

F is the force entered

q is the charge

v is the velocity

B is the magnetic field

Given that,

The magnetic field is

B = 2•i + 4•j. T

The velocity of the electron is

v = 2•i + 6•j + 8•k. m/s

Also, the charge of an electron is

q = -1.602 × 10^-19 C.

Then note that,

V×B is the cross product of the speed and the magnetic field

Then,

F = q (V×B)

F = -1.602 × 10^-19( 2•i + 4•j +8•k × 2•i + 4•j)

Note

i×i=j×j×k×k=0

i×j=k. j×i=-k

j×k=i. k×j=-i

k×i=j. i×k=-j

F = -1.602 × 10^-19[(2•i + 4•j +8•k) × (2•i + 4•j)]

F = -1.602 × 10^-19 [2×2•(i×i) + 2×4•(i×j) + 4×2•(j×i) + 4×4•(j×j) + 8×2•(k×i) + 8×4•(k×j)]

F = -1.602 × 10^-19[4•0 + 8•k + 8•-k + 16•0 + 16•j + 32•-i]

F = -1.602 × 10^-19(0 + 8•k - 8•k + 0 + 16•j - 32•i)

F = -1.602 × 10^-19(16•j - 32•i)

F = -1.602 × 10^-19 × ( -32•i + 16•j)

F = 5.126 × 10^-18 •i - 2.563 × 10^-18 •j

Then, the x component of the force is

Fx = 5.126 × 10^-18 N

Also, the y component of the force is

Fy = -2.563 × 10^-18 N

8 0

3 years ago

A major advantage of case studies is ________.