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3 years ago

All magnetic fields result from the movement of

Physics

1 answer:

konstantin123 [22]3 years ago

5 0

Don’t know sorry I’m just trying not a good person

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A crate is given a big push, and after it is released, it slides up an inclined plane which makes an angle 0.52 radians with the

vovangra [49]

Answer:

a = 8.951 m/s²

Explanation:

given,

angle = 0.52 radians

μ_s = 0.84

μ_k = 0.48

acceleration = ?

using

F + f = m a

mg sin θ + μk mg cos θ = m a

a = g sin θ + μk g cos θ

a = 9.8 x sin 0.52 + 0.48 x 9.8 x cos 0.52

a = 4.869 + 4.082

a = 8.951 m/s²

the magnitude of acceleration is a = 8.951 m/s²

8 0

3 years ago

An empty truck and a fully loaded truck are next to each other at the traffic light. When the light turns green the lighter truc

Lilit [14]

This is correct. The lighter truck moves faster.

4 0

3 years ago

Physics question A spacecraft descends vertically near the surface of Planet X. An upward thrust of 25.0kN from its engines slow

Evgesh-ka [11]

F=m*a and m is constant on any planet
25000-m*g=m*1.2
10000-m*g=-m*0.80
m*g is the weight
25000/1.2-m*g/1.2=m
10000/0.80-m*g/0.80=-25000/1.2+m*g/1.2 solve for m*g
m*g=(10000/0.80+25000/1.2)/ (1/1.2+1/0.80)
16 kN

4 0

3 years ago

The distance between the first and fifth minima of a single-slit diffraction pattern is 0.500 mm with the screen 37.0 cm away fr

WARRIOR [948]

In the single-slit experiment, the displacement of the minima of the diffraction pattern on the screen is given by
$y_n= \frac{n \lambda D}{a}$ (1)
where
n is the order of the minimum
y is the displacement of the nth-minimum from the center of the diffraction pattern
$\lambda$ is the light's wavelength
D is the distance of the screen from the slit
a is the width of the slit

In our problem,
$D=37.0 cm=0.37 m$
$\lambda=530 nm=5.3 \cdot 10^{-7} m$
while the distance between the first and the fifth minima is
$y_5-y_1 = 0.500 mm=0.5 \cdot 10^{-3} m$ (2)

If we use the formula to rewrite $y_5, y_1$ , eq.(2) becomes
$\frac{5 \lambda D}{a} - \frac{1 \lambda D}{a} =\frac{4 \lambda D}{a}= 0.5 \cdot 10^{-3} m$
Which we can solve to find a, the width of the slit:
$a= \frac{4 \lambda D}{0.5 \cdot 10^{-3} m}= \frac{4 (5.3 \cdot 10^{-7} m)(0.37 m)}{0.5 \cdot 10^{-3} m}= 1.57 \cdot 10^{-3} m=1.57 mm$

7 0

3 years ago

An electron, starting from rest and moving with a constant acceleration, travels 2.0 cm in 5.0 ms. What is the magnitude of this

IrinaK [193]

<span>s=2.7 centimeters = 0.027 meters
t=3.9 milliseconds = 0.0039 seconds

s=(1/2)a*t^2
so
a=(2.7*2)/(0.0039)^2
= 355,029.58 m/s^2

a=355,029.58 m/s^2 = 355.02958 km / s^2</span>

7 0

2 years ago