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3 years ago

How does a force do negative work on an object

Physics

2 answers:

Lynna [10]3 years ago

8 0

Answer:

Explanation:

How does a force do negative work on an object ; This is when the direction of motion is opposite to the direction of the force

ohaa [14]3 years ago

5 0

Answer:

The physics definition of "work" is: ... Work can be either positive or negative: if the force has a component in the same direction as the displacement of the object, the force is doing positive work. If the force has a component in the direction opposite to the displacement, the force does negative work

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What happend to the egg in the fresh water glass?

Molodets [167]

Answer:

Nothing in the fresh water

The egg will be salty in salt water

Mark as brainlist

8 0

3 years ago

Read 2 more answers

The electric flux through a spherical surface is 1.4 ✕ 105 N · m2/C. What is the net charge (in C) enclosed by the surface?

Anit [1.1K]

Answer:

The value is $Q_{net} = 1.239 *10^{-6} \ C$

Explanation:

From the question we are told that

The electric flux is $\Phi = 1.4*10^{5} \ N\cdot m^2/C$

Generally the net charge is mathematically represented as

$Q_{net} = \Phi * \epsilon_o$

Here $\epsilon_o$ is the permetivity of free space with value

$\epsilon_o = 8.85*10^{-12} \ \ m^{-3} \cdot kg^{-1}\cdot s^4 \cdot A^2$

$Q_{net} = 1.4*10^5 * 8.85*10^{-12}$

=> $Q_{net} = 1.239 *10^{-6} \ C$

8 0

3 years ago

A) A real object 4.0 cm high stands 30.0 cm in front of a converging lens of focal length 23 cm. Find the image distance, the im

vfiekz [6]

Explanation:

Given that,

Height of object = 4.0 cm

Distance of the object u= -30.0 cm

Focal length = 23 cm

We need to calculate the image distance

Using lens's formula

$\dfrac{1}{v}-\dfrac{1}{u}=\dfrac{1}{f}$

Where, u = object distance

v = image distance

f = focal length

Put the value into the formula

$\dfrac{1}{v}-\dfrac{1}{-30}=\dfrac{1}{23}$

$\dfrac{1}{v}=\dfrac{1}{23}-\dfrac{1}{30}$

$\dfrac{1}{v}=\dfrac{7}{690}$

$v=98.57\ cm$

We need to calculate the height of the image

Using formula of height

$\dfrac{h'}{h}=\dfrac{-v}{u}$

Where, h' = height of image

h = height of object

$\dfrac{h'}{4}=\dfrac{-98.57}{-30}$

$h'=\dfrac{98.57\times4}{30}$

$h'=13.14\ cm$

The image is real and inverted.

(b). Now, object distance u = 13.0 cm

We need to calculate the image distance

Using lens's formula

$\dfrac{1}{v}-\dfrac{1}{u}=\dfrac{1}{f}$

Put the value into the formula

$\dfrac{1}{v}=\dfrac{1}{23}-\dfrac{1}{13.0}$

$\dfrac{1}{v}=-\dfrac{10}{299}$

$v=-29.9\ cm$

We need to calculate the height of the image

$\dfrac{h'}{4}=\dfrac{-(-29.9)}{-13.0}$

$h=-\dfrac{29.9\times4}{13.0}$

$h=-9.2\ cm$

The image is virtual and erect.

Hence, This is required solution.

6 0

3 years ago

A 45.5-turn circular coil of radius 4.85 cm can be oriented in any direction in a uniform magnetic field having a magnitude of 0

il63 [147K]

Answer:

4.399 Nm

Explanation:

The maximum Torque on a coil is given as,

τ = BNIA...................... Equation 1

Where τ = Maximum torque exerted on the coil, B = Magnetic Field, N = Number of turns, I = Current, A = Area.

Given: N = 45.5 Turns, B = 0.49 T, I = 26.7 mA = 0.0267 A,

A = πr², Where r = radius of the coil, r= 4.85 cm = 0.0485 m

A = 3.14(0.0485)²

A = 7.39×10⁻³ m².

Substitute into equation 1

τ = 45.5×0.49×26.7×7.39×10⁻³

τ = 4.399 Nm

3 0

3 years ago

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A level curve on a country road has a radius of 150 m. What is the maximum speed at which this curve can be safely negotiated on

SSSSS [86.1K]

The maximum speed of the car is 24.3 m/s

Explanation:

For a car moving along an unbanked turn, the frictional force provides the centripetal force required to keep the car in circular motion. Therefore, we can write:

$\mu mg = m\frac{v^2}{r}$