What is the relationship between force and momentum?

Mazyrski [523]

6 degrees
4N
.....................

5 0

3 years ago

Take one jar with salt water and another with pure water .put an egg into both jars. In which the jar egg floated over the liqui

DanielleElmas [232]

Answer:

if the which coduct salt has more density so the egg floats .

Explanation:

8 0

3 years ago

What is the colour of a shirt which has blue and white stripes when put under yellow light? Explain.

GenaCL600 [577]

Answer:

Black and yellow strips.

Explanation:

Color of shirt is blue and white stripes

When the shirt is seen in the yellow light, the blue color do not reflect any color so it appears black and the white color reflects the yellow light so it appears yellow.

Thus, the color of shirt is blue appears black and white appears yellow.

8 0

3 years ago

Magnitude of u = 15, direction angle θ = 35° Magnitude of v = 18, direction angle θ = 60° Find the magnitude and direction angle

djverab [1.8K]

Answer:

32.225 and angle is 48.7 degree.

Explanation:

u = 15, θ = 35 degree

v = 18, θ = 60

First represent the u and v in vector form.

u = 15 (Cos 35 i + Sin 35 j ) = 12.287 i + 8.6 j

v = 18 ( Cos 60 i + Sin 60 j ) = 9 i + 15.6 j

The sum of the two vectors is given by

u + v = 12.287 i + 8.6 j + 9 i + 15.6 j = 21.28 i + 24.2 j

Magnitude of u + v = $\sqrt{21.28^{2}+24.2^{2}}$ = 32.225

Let Ф be the angle

tan Ф = 24.2 / 21.28

Ф = 48.7 degree

5 0

3 years ago

At noon, ship A is 110 km west of ship B. Ship A is sailing east at 20 km/h and ship B is sailing north at 15 km/h. How fast is

Katyanochek1 [597]

Answer:

$4.47\ \text{km/h}$

Explanation:

$\dfrac{da}{dt}$ = Rate at which the distance between A and starting point of B is changing = -20 km/h

$\dfrac{db}{dt}$ = Rate at which the distance of B is changing = 15 km/h

$\dfrac{dc}{dt}$ = Rate at which the distance between A and B is changing

Time after which the rate at which the distance between A and B is changing is 4 hours

Distance covered by A in 4 hours = $20\times 4=80\ \text{km}$

a = Distance remaining to the start point of B = $110-80=30\ \text{km}$

b = Distance covered by B in 4 hours = $15\times 4=60\ \text{km}$

Distance between A and B after 4 hours

$c=\sqrt{a^2+b^2}\\\Rightarrow c=\sqrt{30^2+60^2}\\\Rightarrow c=67.08\ \text{km}$

$c^2=a^2+b^2$

Differentiating with respect to time we get

$c\dfrac{dc}{dt}=a\dfrac{da}{dt}+b\dfrac{db}{dt}\\\Rightarrow \dfrac{dc}{dt}=\dfrac{a\dfrac{da}{dt}+b\dfrac{db}{dt}}{c}\\\Rightarrow \dfrac{dc}{dt}=\dfrac{30\times -20+60\times 15}{67.08}\\\Rightarrow \dfrac{dc}{dt}=4.47\ \text{km/h}$

The rate at which the distance between the ships is changing at 4 PM is $4.47\ \text{km/h}$ .

7 0

3 years ago