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faltersainse [42]

4 years ago

15

Select the correct answer

1 answer:

Flauer [41]4 years ago

5 0

Answer: The correct answer is 1800 meters

Explanation: The velocity is equal to distance over time. Thus the formula for distance is d=vt

And 1minute = 60 seconds

d= 30m/s×60s

d= 1800m

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A force is required for any change in an object's motion to occur.<br><br> True<br> False

sertanlavr [38]

Answer:

True

Explanation:

an object in motion stays in motion unless acted upon by another force

7 0

3 years ago

Read 2 more answers

A ray of light has a wavelength of

MArishka [77]

Answer:

The wavelength in vacuum is equal to 428.8 nm.

Explanation:

Given that,

The wavelength of light, $\lambda=284\ nm$

The refractive index of glass, n = 1.51

We need to find the wavelength in vacuum. The relation between wavelength and refractive index is given by :

$n=\dfrac{\lambda_v}{\lambda}\\\\\lambda_v=n\times \lambda\\\\\lambda_v=1.51\times 284\\\\\lambda_v=428.8\ nm$

So, the wavelength in vacuum is equal to 428.8 nm.

6 0

3 years ago

An object is located in air, 25 cm from the vertex of the concave surface of a block of glass (as viewed from the air side of th

Marizza181 [45]

Your question is missing one part as "magnification"

i have completed the missing part below

Answer:

a. $d_{i}=-0.0566cm$

b. $M=441.69$

Explanation:

For this type of numerical we will use the following formulas

$\frac{n1}{d_{o} }+\frac{n_{2} }{d_{i} }=\frac{n_{2}-n_{1} }{R}$ ......... Eq1

where,

$n_{1}$ $=$ refractive index of the medium surrounding refracting surface/object

i.e. $n_{1}$ = $n_{air}$

$n_{2}$ $=$ refractive index of the refracting surface/object

i.e. $n_{2}=n_{glass}$

$d_{0}=$ distance of object from the vertex of the refracting surface

$d_{i}=$ distance of image from the vertex of the refracting surface

$R=$ radius of curvature of the refracting surface

$M=\frac{d_{0} }{d_{i} }$ ........... Eq2

where,

$M=$ magnification

Convention:

$R>0\for\ the\ convex\ refractive\ surface\ of\ curvature\\\ R$

Given:

$n_{1}$ = $n_{air}$ $=1.0$

$n_{2}=n_{glass}$ $=1.5$

$d_{0}=$ $25cm$

$R=-11$ $cm$ because refraction surface is concave

Required:

a. $d_{i}=$ $?$

b. $M=?$

Solution:

a. putting values in eq1, we get

$\frac{1.0}{25}+\frac{1.5}{d_{i} }=\frac{1.5-1.0}{-11}$

$\frac{1.5}{d_{i} }=-0.045-0.040$

$d_{i}=(-0.085)(\frac{1}{1.5} )$

$d_{i}=-0.0566$ cm

b. $M=\frac{25}{-0.05665}$

$M=441.69$

5 0

4 years ago

The weight of spaceman Speff at the surface of planet X, solely due to its gravitational pull, is 389 N. If he moves to a distan

9966 [12]

Answer

given,

W = weight of speff at surface of planet = 384 N

h= height = 1.86 x 10⁴ Km

W' = weight of speff at height h from the planet = 24.31 N

Ms = speff mass = 75 kg

M= mass of planet = ?

R = radius of planet

G = 6.67 x 10⁻¹¹ m³/kg.s²

$g = \dfrac{W}{M}$

$g = \dfrac{389}{75}$

g = 5.187 m/s²

we know,

$g = \dfrac{GM}{r^2}$

$5.19 = \dfrac{GM}{r^2}$ .......(1)

now,

$g' = \dfrac{W}{M}$

$g' = \dfrac{24.31}{75}$

g' = 0.324 m/s²

$g = \dfrac{GM}{(r+h)^2}$

$0.324= \dfrac{GM}{(r+h)^2}$ .........(2)

dividing equation (1)/(2)

$\dfrac{5.19}{0.324}=(\dfrac{(r+h)}{r})^2$

$4=1+\dfrac{h}{r}$

$r = \dfrac{1.86\times 10^4}{3}$

r = 6.2 x 10⁶ m

mass of planet X

from equation 1

$M = \dfrac{g.r^2}{G}$

$M = \dfrac{5.19 \times (6.2\times 10^6)^2}{6.2 \times 10^{-11}}$

M = 3.21 x 10²⁴ kg

8 0

3 years ago

If a material has 3 elements but not joined in a fixed proportion, it is a.

KengaRu [80]

If the proportions are fixed, it's probably a compound, but not necessarily.

If the proportions are NOT fixed, it's a mixture.

The number of components doesn't matter.

3 0

3 years ago