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3 years ago

A box is at rest on a table. What can you say about the forces acting on the box?

2 answers:

6 0

You can tell a lot about an object that's not moving,
and also a lot about the forces acting on it:

==> If the box is at rest on the table, then it is not accelerating.

==> Since it is not accelerating, I can say that the forces on it are balanced.

==> That means that the sum of all forces acting on the box is zero,
and the effect of all the forces acting on it is the same as if there were
no forces acting on it at all.

==> This in turn means that all of the horizontal forces are balanced,
AND all of the vertical forces are balanced.

Horizontal forces:
sliding friction, somebody pushing the box

All of the forces on this list must add up to zero. So ...

(sliding friction force) = (pushing force), in the opposite direction.

If nobody pushing the box, then sliding friction force = zero.

Vertical forces:
gravitational force (weight of the box, pulling it down)
normal force (table pushing the box up)

All of the forces on this list must add up to zero, so ...

(Gravitational force down) + (normal force up) = zero

(Gravitational force down) = -(normal force up) .

Troyanec [42]3 years ago

4 0

1) normal force 2) gravitational force

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A long, straight, horizontal wire carries a left-to-right current of 40 A. If the wire is placed in a uniform magnetic field of

Drupady [299]

Answer:

$4.5\times 10^{-5} T$

Explanation:

We are given that

Current in wire=40 A

Magnetic field= $B_1=3.5\times 10^{-5}$ T( vertically downward)

We have to find the resultant magnitude of the magnetic field 29 cm above the wire and 29 cm below the wire.

According to Bio-Savart law, the magnetic field exerted by the wire at distance R is given by

$B_{wire}=B_2=\frac{\mu_0I}{2\pi R}$

We have R=29 cm= $\frac{29}{100}=0.29 m$

1 m=100 cm

Substitute the values in the given formula

$B_2=\frac{4\pi\times 10^{-7}\times 40}{2\times \pi\times 0.29}=\frac{2\times 40\times 10^{-7}}{0.29}=2.76\times 10^{-5} T$

The resultant magnetic field is given by

$B=\sqrt{B^2_1+B^2_2}$

Substitute the values then we get

$B=\sqrt{(3.5\times 10^{-5})^2+(2.76\times 10^{-5})^2}$

$B=4.5\times 10^{-5} T$

The resultant magnitude of magnetic field is same above and below the wire as it is at same distance.

The resultant magnitude of the magnetic field 29 cm below the wire= $4.5\times 10^{-5} T$

Hence, the resultant magnitude of the magnetic field 29 cm above the wire= $4.5\times 10^{-5} T$

7 0

3 years ago

Protons are the smallest particle of an element that retains the original Characteristics of the element. True or False

dlinn [17]

Answer:

False; that’s an atom

Explanation:

5 0

3 years ago

Read 2 more answers

One hazard of space travel is the debris left by previous missions. There are several thousand objects orbiting Earth that are l

MariettaO [177]

Answer:

F = 6666.7 N

Explanation:

Given that,

Mass of a chip, m = 0.1 mg

Initial speed, u = 0

Final speed, $v=4\times 10^{3}\ m/s$

Time of collision, $t=6\times 10^{-8}\ s$

We know that,

Force, F = ma

Put all the values,

$F=\dfrac{m(v-u)}{t}\\\\F=\dfrac{0.1\times 10^{-6}\times (4\times 10^3-0)}{6\times 10^{-8}}\\\\F=6666.7\ N$

So, the required force is 6666.7 N.

3 0

3 years ago

The even distribution of weight in a compostion is known as

mojhsa [17]

The answer is Balance. hopei helped? ahahahaha

3 0

3 years ago

A particular spiral galaxy can be approximated by a thin disk-like volume 62 Thousand Light Years in radius and 7 Hundred Light

SVETLANKA909090 [29]

Answer:

Approximate linear dimension is 2 light years.

Explanation:

Radius of the spiral galaxy r = 62000 LY

Thickness of the galaxy h = 700 LY

Volume of the galaxy = πr²h

= (3.14)(62000)²(700)

= (3.14)(62)²(7)(10)⁸

= 84568×10⁸

= $8.45\times 10^{12}$ (LY)³

Since galaxy contains number of stars = 1078 billion stars ≈ $1.078\times 10^{12}$

Now volume covered by each star of the galaxy = $\frac{\text{Total volume of the galaxy}}{\text{Number of stars}}$

= $\frac{8.45\times 10^{12} }{1.078\times 10^{12}}$

= 7.839 Light Years

Now the linear dimension across the volume

= $(\text{Average volume per star})^{\frac{1}{3}}$

= $(7.839)^{\frac{1}{3}}$

= 1.99 LY

≈ 2 Light Years

Therefore, approximate linear dimension is 2 light years.

8 0

2 years ago