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3 years ago

Marcus drove his Honda Prelude for 4 hours at a rate of 55 miles per hour. How far did he travel?

2 answers:

7 0

To solve this, you’d multiply 55 by 4, because he is travelling 55 miles every hour, for four hours, which means 55 miles every hour. The answer would be 220.

lilavasa [31]3 years ago

5 0

Answer : Distance covered is 220 miles.

Explanation :

It is given that,

Speed of Marcus's Honda Prelude, $v=55\ miles/hour$

Time duration, $t=4\ hours$

We know that the speed of an object is defined as the distance travelled per unit time.

$v=\dfrac{d}{t}$

$d=v\times t$

$d=55\ miles/hour\times 4\ hours$

$d=220\ miles$

Hence, this is the required solution.

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An object kept in front of lens, which is thick at the edges, produces a _________,

Wewaii [24]

Answer:

Explanation:

A diverging lens forms a virtual image that is as the same size as the object

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2 years ago

Problem One: A beam of red light (656 nm) enters from air into the side of a glass and then into water. wavelength, c. and speed

Ivanshal [37]

Answer:

Part a)

$f_w = f_g = 4.57 \times 10^{14} Hz$

Part b)

$\lambda_w = 492 nm$

$\lambda_g = 437.3 nm$

Part c)

$v_w = 2.25 \times 10^8 m/s$

$v_g = 2.0 \times 10^8 m/s$

Explanation:

Part a)

frequency of light will not change with change in medium but it will depend on the source only

so here frequency of light will remain same in both water and glass and it will be same as that in air

$f = \frac{v}{\lambda}$

$f = \frac{3 \times 10^8}{656 \times 10^{-9}}$

$f = 4.57 \times 10^{14} Hz$

Part b)

As we know that the refractive index of water is given as

$\mu_w = 4/3$

so the wavelength in the water medium is given as

$\lambda_w = \frac{\lambda}{\mu_w}$

$\lambda_w = \frac{656 nm}{4/3}$

$\lambda_w = 492 nm$

Similarly the refractive index of glass is given as

$\mu_w = 3/2$

so the wavelength in the glass medium is given as

$\lambda_g = \frac{\lambda}{\mu_g}$

$\lambda_g = \frac{656 nm}{3/2}$

$\lambda_g = 437.3 nm$

Part c)

Speed of the wave in water is given as

$v_w = \frac{c}{\mu_w}$

$v_w = \frac{3 \times 10^8}{4/3}$

$v_w = 2.25 \times 10^8 m/s$

Speed of the wave in glass is given as

$v_g = \frac{c}{\mu_g}$

$v_g = \frac{3 \times 10^8}{3/2}$

$v_g = 2 \times 10^8 m/s$

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3 years ago

Why are negative ions usually larger than positive ions

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2 years ago

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A "biconvex" lens is one in which both surfaces of the lens bulge outwards. Suppose you had a biconvex lens with radii of curvat

Stolb23 [73]

Answer: f=150cm in water and f=60cm in air.

Explanation: Focal length is a measurement of how strong light is converged or diverged by a system. To find the variable, it can be used the formula:

$\frac{1}{f}$ = (nglass - ni)( $\frac{1}{R1}$ - $\frac{1}{R2}$ ).

nglass is the index of refraction of the glass;

ni is the index of refraction of the medium you want, water in this case;

R1 is the curvature through which light enters the lens;

R2 is the curvature of the surface which it exits the lens;

Substituting and calculating for water (nwater = 1.3):

$\frac{1}{f}$ = (1.5 - 1.3)( $\frac{1}{10}$ - $\frac{1}{15}$ )

$\frac{1}{f}$ = 0.2( $\frac{1}{30}$ )

f = $\frac{30}{0.2}$ = 150

For air (nair = 1):

$\frac{1}{f}$ = (1.5 - 1)( $\frac{1}{10}$ - $\frac{1}{15}$ )

f = $\frac{30}{0.5}$ = 60

In water, the focal length of the lens is f = 150cm.

In air, f = 60cm.

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3 years ago

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