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2 years ago

This is my question. I need the answer stat.

Physics

2 answers:

Delvig [45]2 years ago

6 0

The safety stuff should be put on before you do anything else.

Even before you load the lab coffeepot.

Kipish [7]2 years ago

5 0

Answer:

Explanatioyour answers look right, but if there has , has to be another answer its a , but your answers are right

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Pani-rosa [81]

The total displacement is 4.0 m east.

4 0

3 years ago

A ball is tossed from an upper-story window of a building. the ball is given an initial velocity of 8.00 m/s at an angle of 20.0

otez555 [7]

A) The motion of the ball consists of two indipendent motions on the horizontal (x) and vertical (y) axis. The laws of motion in the two directions are:
$x(t)=v_0 \cos \alpha t$
$y(t)=h-v_0 \sin \alpha t - \frac{1}{2}gt^2$
where
- the horizontal motion is a uniform motion, with constant speed $v_0 \cos \alpha$ , where $v_0 = 8.00 m/s$ and $\alpha=20.0^{\circ}$
- the vertical motion is an uniformly accelerated motion, with constant acceleration $g=9.81 m/s^2$ , initial position h (the height of the building) and initial vertical velocity $v_0 \sin \alpha$ (with a negative sign, since it points downwards)

The ball strikes the ground after a time t=3.00 s, so we can find the distance covered horizontally by the ball by substituting t=3.00 s into the equation of x(t):
$x(3.00 s)=v_0 \cos \alpha t=(8 m/s)(\cos 20^{\circ})(3.0 s)=22.6 m$

b) To find the height from which the ball was thrown, h, we must substitute t=3.00 s into the equation of y(t), and requiring that y(3.00 s)=0 (in fact, after 3 seconds the ball reaches the ground, so its vertical position y(t) is zero). Therefore, we have:
$0=h-v_0 \sin \alpha t - \frac{1}{2}gt^2$
which becomes
$h=(8 m/s)(\sin 20^{\circ})(3.0 s)+ \frac{1}{2}(9.81 m/s^2)(3.0 s)^2=52.3 m$

c) We want the ball to reach a point 10.0 meters below the level of launching, so we want to find the time t such that
$y(t)=h-10$
If we substitute this into the equation of y(t), we have
$h-10 = h-v_0 \sin \alpha t- \frac{1}{2}gt^2$
$\frac{1}{2}gt^2+v_0 \sin \alpha t -10 =0$
$4.9 t^2 +2.74 t-10 =0$
whose solution is $t=1.18 s$ (the other solution is negative, so it has no physical meaning). Therefore, the ball reaches a point 10 meters below the level of launching after 1.18 s.

4 0

4 years ago

A cube, whose edges are aligned with the , and axes, has a side length . The field is immersed in an electric field aligned with

Ivenika [448]

Complete Question

Complete Question is attached below

Answer:

$q=1.558*10^{-9}c$

Explanation:

From the question we are told that:

Side length s=1.13m

Left field strength $E_l=784.75N/m$

Right field strength $E_r=776.38 N/m$

Front field strength $E_f=725.5 N/m$

Back field strength $E_b=749.54 N/m$

Top field strength $E_t=944.95 N/m$

Bottom field strength $E_{bo}=1082.58 N/m$

Generally, the equation for Charge flux is mathematically given by

$\phi=EAcos\theta$

Where

Theta for Right,Left,Front and Back are at an angle 90

$cos 90=0$

Therefore

$\phi =0$ with respect to Right,Left,Front and Back

Generally, the equation for Charge Flux is mathematically also given by

$\phi=\frac{q}{e_o}$

Where

$Area =L*B\\\\A=1.13*1.13\\\\A=1.2769m^2$

Therefore

$Q_{net}=E_{bo}Acos\theta_{bo} +E_tAcos\theta_t$

$Q_{net}=1082.85*1.2769*cos0=944.95*1.2769cos (180)$

$Q_{net}=176N/C m^2$

Giving

$q=\phi*e_0$

$q=176N/C m^2*1.558*10^{-12}c$

$q=1.558*10^{-9}c$

4 0

3 years ago

How do I find an apparent weight in N for a metal connected to a string submerged in water if a scale shows the mass 29.52 g whe

Rudiy27

The Tension of the string is going to be less when submerged in water by a value called the buoyancy force, so below in the attached file is explanation on how to calculate the apparent weight and density of the submerged object