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3 years ago

15

A block starting from rest slides down the length of an 18 plank with an acceleration of 4.0 meters per second. How long does th

e block take to reach the bottom?

1 answer:

Snowcat [4.5K]3 years ago

4 0

Answer:

$\boxed{\text{\sf \Large 3.0 s}}$

Explanation:

Use distance formula

$\displaystyle d=ut+\frac{1}{2} at^2$

$u= \text{\sf initial velocity}\\d= \text{\sf distance}\\a= \text{\sf acceleration}\\t= \text{\sf time taken}$

$\displaystyle 18=0 \times t+\frac{1}{2} \times 4 \times t^2$

$t=3$

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A meteor moving 468 km per minute traveling in a south-to-north direction passed near Earth in 2013. Does this statement describ

Thepotemich [5.8K]

Answer:

this statement describes meteor's velocity,

because velocity is a vector quantity which has both magnitude as well as a specific direction and here the meteor's direction is specified in the statement hence we conclude that this statement describes meteor's velocity as well as speed too.

3 0

3 years ago

A short current element dl⃗ = (0.500 mm)j^ carries a current of 5.40 A in the same direction as dl⃗ . Point P is located at r⃗ =

SashulF [63]

Answer:

The magnetic field along x axis is

$B_{x}=1.670\times10^{-10}\ T$

The magnetic field along y axis is zero.

The magnetic field along z axis is

$B_{z}=3.484\times10^{-10}\ T$

Explanation:

Given that,

Length of the current element $dl=(0.5\times10^{-3})j$

Current in y direction = 5.40 A

Point P located at $\vec{r}=(-0.730)i+(0.390)k$

The distance is

$|\vec{r}|=\sqrt{(0.730)^2+(0.390)^2}$

$|\vec{r}|=0.827\ m$

We need to calculate the magnetic field

Using Biot-savart law

$B=\dfrac{\mu_{0}}{4\pi}\timesI\times\dfrac{\vec{dl}\times\vec{r}}{|\vec{r}|^3}$

Put the value into the formula

$B=10^{-7}\times5.40\times\dfrac{(0.5\times10^{-3})\times(-0.730)i+(0.390)k}{(0.827)^3}$

We need to calculate the value of $\vec{dl}\times\vec{r}$

$\vec{dl}\times\vec{r}=(0.5\times10^{-3})\times(-0.730)i+(0.390)k$

$\vec{dl}\times\vec{r}=i(0.350\times0.5\times10^{-3}-0)+k(0+0.730\times0.5\times10^{-3})$

$\vec{dl}\times\vec{r}=0.000175i+0.000365k$

Put the value into the formula of magnetic field

$B=10^{-7}\times5.40\times\dfrac{(0.000175i+0.000365k)}{(0.827)^3}$

$B=1.670\times10^{-10}i+3.484\times10^{-10}k$

Hence, The magnetic field along x axis is

$B_{x}=1.670\times10^{-10}\ T$

The magnetic field along y axis is zero.

The magnetic field along z axis is

$B_{z}=3.484\times10^{-10}\ T$

7 0

3 years ago

A thin ring of radius 73 cm carries a positive charge of 610 nC uniformly distributed over it. A point charge q is placed at the

kow [346]

Answer:

q = - 93.334 nC

Explanation:

GIVEN DATA:

Radius of ring 73 cm

charge on ring 610 nC

ELECTRIC FIELD p FROM CENTRE IS AT 70 CM

E = 2000 N/C

Electric field due tor ring is guiven as

$E = \frac{KQx}{[x^2+ R^2]^{3/2}}$

$E = \frac{9\time 10^9 \times 610\times 10^[-9} 0.70}{(0.70^2 + 0.73^2)^{3/2}}$

E1 = 3714.672 N/C

electric field due to point charge q

$E =\frac[kq}{x^2}$

$E = \frac{9\times 10^9 \times q}{0.70^2}$

$E2 = 1.837\times 10^{10}\times q$

now the eelctric charge at point P is

E = E1 + E2 $2000 = 3714.672 + 1.837\times 10[10} \times q$

solving for q

q = - 93.334 nC

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3 years ago

Which of the following statements describe the steps you need to take to delete a file?

Doss [256]

In order to delete a file, it is required to first select the file which you want to delete and then select 'delete' from the file menu. It will then ask you to confirm the deletion which will be required in order to complete the file deletion process.

It isn't required to click on the X at the top right of the screen or to open the file you want to delete as these steps aren't relevant to the deletion process.

Hence, the statements which describe the steps you need to take in order to delete a file are as follows:

- Confirm the deletion.
- Select Delete from the File menu.
- Select the file you want to delete.

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2 years ago

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Which is a property of light a: mass b: density c: wavelength d: pitch

Reptile [31]

Im pretty sure its C- Wavelength

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3 years ago