Given a force of 100 N and an acceleration of 10 m/s2, what is the mass?

A school bus takes 0.690 h to reach the school from your house. If the average velocity of the bus is 18.7 km/h to the East, wha

Dennis_Churaev [7]

The displacement is 12.903 km east.

4 0

3 years ago

An athlete ran a distance of 100 meters in 9.83 Seconds<br> What’s his average speed

trasher [3.6K]

Answer:

10.1729399797mp/s

Explanation:

100m divided by 9.83s=

10.1729399797mp/s

6 0

3 years ago

A car is driven 13 mi east and then certain distance due north, ending up at the position 25 degrees north of east of its initia

Kobotan [32]

Answer: Hello mate!

lets define the north as the y-axis and east as the x-axis.

Using the notation (x,y) we can define the initial position of the car as (0,0)

then the car travells 13 mi east, so now the position is (13,0)

then the car travels Y miles to the north, so the position now is (13, Y)

and we know that the final position is 25° degrees north of east of the initial position. This angle says that the distance traveled to the north is less than 13 mi because this angle is closer to the x-axis (or east in this case).

This angle is measured from east to north, then the adjacent cathetus is on the x-axis, in this case, 13mi

And we want to find the distance Y, so we can use the tangent:

Tan(25°) = Y/13

tan(25°)*13 mi = Y = 6.06 mi.

3 0

3 years ago

The apparent depth of an object at the bottom of a tank filled with a liquid of refractive index 1.3 is 7.7 cm. What is the actu

Juliette [100K]

$\orange{\underline{\huge{\bold{\textit{\green{\bf{QUESTION}}}}}}}$

The apparent depth of an object at the bottom of a tank filled with a liquid of refractive index 1.3 is 7.7 cm. What is the actual depth of the liquid in the tank?

${\bold{\blue{GIVEN}}}$

REFRACTIVE INDEX = 1.3

APPARENT DEPTH = 7.7 cm

${ \bold{\green{To \: Find}}}$

REAL DEPTH OF THE OBJECT.

${\red{FORMULA \: \: USED }}$

$Reflective \: Index = \frac{Real \: Depth}{Apparent \: Depth }$

$\huge\mathbb{\red A \pink{N}\purple{S} \blue{W} \orange{ER}}$

Refractive Index = 1.3

Apparent Depth = 7.7 cm

Putting the values in the formula:-

$Reflective \: Index = \frac{Real \: Depth}{Apparent \: Depth } \\ \\ 1.3 = \frac{Real \: Depth}{7.7 \: cm} \\ \\ 1.3 \times 7.7 = Real \: Depth \\ \\ 10.01 \: \: cm = Real \: Depth$