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2 years ago

How is a crowbar used to remove the top of a box change the direction of the force

1 answer:

7 0

The crowbar is placed under a small edge under the top of the box. The crowbar will send the box lid flying from the opposite direction that the crowbar is placed. Please give Brainlest if you understand.

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What is the mass in rams of 30.00 ml of a liquid that has a density of 1.21 g/ml?

Murljashka [212]

Answer:

36.3 g

Explanation:

1.21 g/ml * 30 ml = 36.3 g ( see how the 'ml' units cancel ?)

3 0

1 year ago

One wire possesses a solid core of copper, with a circular cross-section of radius 3.78 mm. The other wire is composed of 19 str

Phoenix [80]

Answer:

a) Current in wire 1 = 0.132 A

Current in wire 2 = 0.101 A

b) Resistance of wire 1 = R₁ = 0.000376 Ω = (3.76 × 10⁻⁴) Ω = 0.376 mΩ

Resistance of wire 2 = R₂ = 0.000495 Ω = (4.95 × 10⁻⁴) Ω = 0.495 mΩ

Explanation:

Current density, J = (current) × (cross sectional area)

Current density for both wires = J = 2950 A/m²

For wire 1,

Cross sectional Area = πr² = π(0.00378²)

A₁ = 0.00004491 m²

For wire 2,

With the assumption that the strands are well banded together with no spaces in btw.

Cross sectional Area = 19 × πr² = π(0.000756)²

A₂ = 0.00003413 m²

Current in wire 1 = I₁ = J × A₁ = 2950 × 0.00004491 = 0.132 A

Current in wire 2 = I₂ = J × A₂ = 2950 × 0.00003413 = 0.101 A

b) Resistance = ρL/A

ρ = resistivity for both wires = (1.69 x 10⁻⁸) Ω.m

L = length of wire = 1.00 m for each of the two wires

A₁ = 0.00004491 m²

A₂ = 0.00003413 m²

R₁ = ρL/A₁ = (1.69 x 10⁻⁸ × 1)/0.00004491

R₁ = 0.000376 Ω = (3.76 × 10⁻⁴) Ω = 0.376 mΩ

R₂ = ρL/A₂ = (1.69 x 10⁻⁸ × 1)/0.00003413

R₂ = 0.000495 Ω = (4.95 × 10⁻⁴) Ω = 0.495 mΩ

Hope this helps!!

4 0

3 years ago

A section of uniform pipe is bent into an upright U shape and partially filled with water, which can then oscillate back and for

nadezda [96]

Answer:

The expression is shown in the picture below

Explanation:

7 0

2 years ago

physics major is cooking breakfast when he notices that the frictional force between the steel spatula and the Dry Steel frying

givi [52]

Answer:0.667 N

Explanation:

Given

It is noticed that Frictional Force is 0.2 N

Coefficient of kinetic Friction $\mu _k=0.3$

We know Friction Force is given by

$f_r=\mu _k\times Normal\ reaction$

therefore

$0.2=\mu _k\times N$

$0.2=0.3\cdot N$

$N=\frac{0.2}{0.3}$

$N=0.667 N$

8 0

3 years ago

A daring 510N swimmersdives off a cliff with a running horizontal lead.what must be her mimimum speed just as she leaves the top

goldfiish [28.3K]

Answer:

The running horizontal speed should be larger than 1.29 m/s.

Explanation:

In order for the swimmer to just miss the bone-breaking ledge, her horizontal speed must be

$v > \frac{1.75m}{t_{fall}}$

in which we need to know how long we she be diving through the air. To determine that, recall the formula for the distance made by an object with acceleration (in this case it is the gravitational acceleration) with no initial (vertical) velocity:

$s = \frac{1}{2}gt^2$

from which it follows that (for non-negative t)

$t = \sqrt{\frac{2\cdot9m}{9.8\fram{m}{s^2}}}\approx1.36s$

This result can be used in the initial inequality:

$v > \frac{1.75m}{t_{fall}}=\frac{1.75m}{1.36s}=1.29\frac{m}{s}$

The diving lady better gets a speed larger than 1.29 m/s to avoid landing on the ledge.

8 0

3 years ago

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