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1 year ago

The force required to maintain an object at a constant velocity in free space is equal to what ?

Physics

1 answer:

hram777 [196]1 year ago

7 0

Answer:

The force required to maintain an object at a constant velocity in free space is equal to Zero.

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How much energy is needed to raise a 50kg block up from the ground to a height of 5 meters?

miss Akunina [59]

Answer:

The answer to your question is Pe = 2452.5 J

Explanation:

Data

mass = 50 kg

height = 5 m

gravity = 9.81 m/s²

Process

The energy of this process is Potential energy which is proportional to the mass of the body, the gravity and the height of the body.

Pe = mgh

Substitution

Pe = (50)(5)(9.81)

Simplification

Pe = 2452.5 J

8 0

2 years ago

A pulley has a mechanical advantage of 1.

sertanlavr [38]

Answer:

There is no mechanical advantage

Explanation:

The mechanical advantage is possible only when the force needed to lift a load is lesser than the weight of the load.

For example, is we have a mechanical advantage of 2, the force needed to lift will be 1/2 of the weight of the load, and if we have a mechanical advantage of 4, the force needed will be 1/4 of the weight of the load.

In the attached image there are clear examples of mechanical advantage with pulleys.

7 0

2 years ago

A large insulating sheet has a surface charge density σ. what is the electric field strength near the insulating sheet?

MakcuM [25]

Free insulative sheets and insulative sheets backed by a grounded conductor are the only cases in which it is possible to extract reliable information from a noncontacting measurement of the charged state of an insulator. In both cases, the electric field from the charge is the deciding factor.<span>

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3 0

2 years ago

A small current element carrying a current of I = 1.00 A is placed at the origin given by d → l = 4.00 m m ^ j Find the magnetic

xxTIMURxx [149]

Answer:

the magnitude and direction of d → B on the x ‑axis at x = 2.50 m is -6.4 × 10⁻¹¹T(Along z direction)

the magnitude and direction of d → B on the z ‑axis at z = 5.00 m is 1.6 × 10⁻¹¹T(Along x direction)

Explanation:

Use Biot, Savart, the magnetic field

$d\bar{B}=\frac{U}{4\pi } \frac{i(d\bar{l}\times r)}{r^2}$

Given that,

i = 1.00A

d → l = 4.00 m m ^ j

r = 2.5m

Displacement vector is

$\bar{r}=x\hat i+y\hat j+z \hat k\\$

$\bar{r}= (2.5m) \hat i +(0m)^2 + (0m)^2$

=2.5m

on the axis of x at x = 2.5

$r = \sqrt{(2.5)^2 + (0)^2 + (0)^2}$

r = 2.5m

And unit vector

$\hat r =\frac{\bar{r}}{r}$

$= \frac{2.5 \hat i}{2.5}\\\\= 1\hat i$

Therefore, the magnetic field is as follow

$d\bar{B}=\frac{U}{4\pi } \frac{i(d\bar{l}\times r)}{r^2}$

$d\bar{B} = \frac{(10^-^7)(1)(4\times10^-^3j\times i}{(2.50)^2} \\\\d\bar{B} = -6.4\times10^{-11} T$

(Along z direction)

B)r = 5.00m

Displacement vector is

$\bar{r}=x\hat i+y\hat j+z \hat k\\$

$\bar{r}= (5.00m) \hat i +(0m)^2 + (0m)^2$

=5.00m

on the axis of x at x = 5.0

$r = \sqrt{(5.00)^2 + (0)^2 + (0)^2}$

r = 5.00m

And unit vector

$\hat r =\frac{\bar{r}}{r}$

$= \frac{5.00 \hat i}{5.00}\\\\= 1\hat i\\$

Therefore, the magnetic field is as follow

$d\bar{B}=\frac{U}{4\pi } \frac{i(d\bar{l}\times r)}{r^2}$

$d\bar{B} = \frac{(10^-^7)(1)(4\times10^-^3j\times i}{(5.00)^2} \\\\d\bar{B} = 1.6\times10^{-11} T$

(Along x direction)

7 0

3 years ago

A grating with 250 grooves mm is used with an incandescent light source. Assume the visible spectrum to range in wavelength from

NeTakaya

The number of orders that one can see the entire visible speed is 5 orders.

<h3>How to calculate the orders?</h3>

From the information given, it should be noted that the grating spacing will be:

d = (1.00 × 10^-4) / 250

d = 4000nm

Therefore, the number of times that are needed to complete the order will be the same as the number of orders which the long wavelength time will be visible. This will be:

= (4000 × sin 90°)/700

= 5.71

Therefore, the maximum orders will be 5.

Learn more about speed on:

brainly.com/question/13943409

#SPJ1

7 0

1 year ago