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3 years ago

~!Here's the question!~

Physics

2 answers:

vazorg [7]3 years ago

8 0

The correct answer is A) Gravity. Hope this helps.

gtnhenbr [62]3 years ago

6 0

The answer would most likely be A since obviously gravity weighs things down which helps the every other masses stay settled in place

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Two long straight wires are parallel and 8.6 cm apart. They are to carry equal currents such that the magnetic field at a point

Neko [114]

Answer:

(a) The current should be in opposite direction

(b) The current needed is 39.8 A

Explanation:

Part (a)

Based, on right hand rule, the current should be in opposite direction

Part (b)

given;

strength of magnetic field, B = 370 µT

distance between the two parallel wires, d = 8.6 cm

$B = \frac{\mu_oI}{2\pi R}$

At the center, the magnetic field strength is twice

$B_c = 2(\frac{\mu_oI}{2\pi R}) =\frac{ \mu_oI}{\pi R}$

R = d/2 = 8.6/2 = 4.3 cm = 0.043 m

$B_c = \frac{ \mu_oI}{\pi R}\\\\I = \frac{B_c\pi R}{\mu_o} = \frac{370 *10^{-6}* \pi *0.043}{4\pi *10^{-7}}\\\\I = 39.8 \ A$

Therefore, current needed is 39.8 A

6 0

2 years ago

Consider horizontal parallel plates with a fixed potential difference. The upper plate has a voltage difference of 30 V with the

BlackZzzverrR [31]

relation between potential difference and electric field is given as

$E . d = \Delta V$

so here we know that

d = 3 cm

$\Delta V = 30 V$

$E \times 0.03 = 30$

$E = 1000 N/C$

So now when plates are separated to 4 cm distance carefully

the potential difference between them will change but the electric field between them will remain constant

So at distance of 4 cm also the electric field will be E = 1000 N/C

5 0

3 years ago

In a shipping company distribution center, an open cart of mass 50.0 kg is rolling to the left at a speed of 5.00 m/s. Ignore fr

spin [16.1K]

Answer:

a) $v_p=9.35m/s$

Explanation:

From the question we are told that:

Open cart of mass $M_o=50.0 kg$

Speed of cart $V=5.00m/s$

Mass of package $M_p=15.0kg$

Speed of package at end of chute $V_c=3.00m/s$

Angle of inclination $\angle =37$

Distance of chute from bottom of cart $d_x=4.00m$

Generally the equation for work energy theory is mathematically given by

$\frac{1}{2}mu^2+mgh=\frac{1}{2}mv_p^2$

Therefore

$\frac{1}{2}u^2+gh=\frac{1}{2}v_p^2$

$v_p=\sqrt{2(\frac{1}{2}u^2+gh)}$

$v_p=\sqrt{2(\frac{1}{2}v_c^2+gd_x)}$

$v_p=\sqrt{2(\frac{1}{2}(3)^2+(9.8)(4))}$

$v_p=9.35m/s$

4 0

2 years ago

What is the gravity force between two stars with mass of 5,000,000 kg and 1,000,000 kg if the distance between them is 100 m

Luda [366]

Answer:

0.0334N

Explanation:

Given parameters:

M1 = 5 x 10⁶kg

M2 = 1 x 10⁶kg

Distance = 100m

Unknown:

Gravitational force = ?

Solution:

To solve this problem, we use the Newton's law of universal gravitation.

Fg = $\frac{G m1 m2}{r^{2} }$

G is the universal gravitation constant

m is the mass

r is the distance

Fg = $\frac{6.67 x 10^{-11} x 5 x 10^{6} x 1 x 10^{6} }{100^{2} }$ = 0.0334N

8 0

3 years ago

An object is dropped from the top of a tall building. At 2 seconds, it is 64 feet from the top of the building. At 4 seconds, it

jeka94

Answer:

96.21 ft/s

Explanation:

To solve this, you only need to use one expression which is:

Vf² = Vo² + 2gh

g = 9.8 m/s²

However, this exercise is talking in feet, so convert the gravity to feet first:

g = 9.8 * 3.28 = 32.15 ft/s²

Vo is zero, because it's a free fall and in free fall the innitial speed is always zero. With this, let's calculate the speed at 2 seconds, with a height of 64 ft, and then with the 256 ft:

V1 = √2*32.15*64

V1 = 64.15 ft/s

V2 = √2*32.15*256

V2 = 128.3 ft/s

So the average rate is:

V = 128.3 + 64.15 / 2

V = 96.22 ft/s

6 0

3 years ago

~~~!Here's the question!~~~

~!Here's the question!~