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2 years ago

A student walks 3

Physics

1 answer:

pochemuha2 years ago

8 0

Answer: The speed of the student is 5.45km/hr.

Explanation: The important concept here is to convert 33 minutes to hours by diving by 60 since one hour is equal to 60 minutes to get 0.55hr

Then velocity is equal to distance over time.

v=d/t

v= 3km/0.55hr

v=5.45km/hr

You might be interested in

The line that is drawn perpendicular to the point at which a wave intersects a boundary is know as the

fomenos

The line that is drawn perpendicular to the point at which a wave intersects a boundary is know as the Normal .

When the normal is drawn, the incident ray makes an angle with it known as the angle of incidence and the reflected ray makes an angle with it known as the angle of incidence. These angles are always equal.

The refracted ray makes an angle with the normal known as angle of refraction. The sin of angle of incidence to the sin of angle of refraction is called the refractive index( μ= sin i / sin r) .

hope all of it helps you!

5 0

2 years ago

Find the moments of inertia Ix, Iy, I0 for a lamina that occupies the part of the disk x2 y2 ≤ 36 in the first quadrant if the d

Tasya [4]

Answer:

I(x) = 1444×k × ${\pi}$

I(y) = 1444×k × ${\pi}$

I(o) = 3888×k × ${\pi}$

Explanation:

Given data

function = x^2 + y^2 ≤ 36

function = x^2 + y^2 ≤ 6^2

to find out

the moments of inertia Ix, Iy, Io

solution

first we consider the polar coordinate (a,θ)

and polar is directly proportional to a²

so p = k × a²

so that

x = a cosθ

y = a sinθ

dA = adθda

I(x) = ∫y²pdA

take limit 0 to 6 for a and o to $\pi /2$ for θ

I(x) = $\int_{0}^{6}\int_{0}^{\pi/2}$ y²p dA

I(x) = $\int_{0}^{6}\int_{0}^{\pi/2}$ (a sinθ)²(k × a²) adθda

I(x) = k $\int_{0}^{6}a^(5)$ da × $\int_{0}^{\pi/2}$ (sin²θ)dθ

I(x) = k $\int_{0}^{6}a^(5)$ da × $\int_{0}^{\pi/2}$ (1-cos2θ)/2 dθ

I(x) = k $({r}^{6}/6)^(5)_0$ × ${θ/2 - sin2θ/4}^{\pi /2}_0$

I(x) = k × $({6}^{6}/6)$ × ( ${\pi /4} - sin\pi /4)$

I(x) = k × $({6}^{5})$ × ${\pi /4}$

I(x) = 1444×k × ${\pi}$ .....................1

and we can say I(x) = I(y) by the symmetry rule

and here I(o) will be I(x) + I(y) i.e

I(o) = 2 × 1444×k × ${\pi}$

I(o) = 3888×k × ${\pi}$ ......................2

3 0

3 years ago

a roller coaster begins at the top of a hill if it acelerates at the rate of 2 m/s2 and has a mass of 2000 kg what net force is

AlladinOne [14]

Answer:

$\boxed {\boxed {\sf 4000 \ Newtons }}$

Explanation:

Force can be found by multiplying the mass by the acceleration.

$F=m*a$

The mass of the roller coaster is 2000 kilograms and the acceleration is 2 meters per second squared.

$m= 2000 \ kg \\a= 2 \ m/s^2$

Substitute the values into the formula.

$F= 2000 \ kg * 2 \ m/s^2$

Multiply.

$F= 4000 \ kg*m/s^2$

1 kg*m/s² is equal to 1 N
Therefore our answer of 4000 kg*m/s² is equal to 4000 Newtons

$F= 4000 \ N$

The net force acting on the roller coaster is 4000 Newtons.

7 0

2 years ago

Two identical trucks have mass 5100 kg when empty, and the maximum permissible load for each is 8000 kg. the first truck, carryi

Oksanka [162]

The 2nd truck was overloaded with a load of 16833 kg instead of the permissible load of 8000 kg. The key here is the conservation of momentum. For the first truck, the momentum is 0(5100 + 4300) The second truck has a starting momentum of 60(5100 + x) And finally, after the collision, the momentum of the whole system is 42(5100 + 4300 + 5100 + x) So let's set the equations for before and after the collision equal to each other. 0(5100 + 4300) + 60(5100 + x) = 42(5100 + 4300 + 5100 + x) And solve for x, first by adding the constant terms 0(5100 + 4300) + 60(5100 + x) = 42(14500 + x) Getting rid of the zero term 60(5100 + x) = 42(14500 + x) Distribute the 60 and the 42. 60*5100 + 60x = 42*14500 + 42x 306000 + 60x = 609000 + 42x Subtract 42x from both sides 306000 + 18x = 609000 Subtract 306000 from both sides 18x = 303000 And divide both sides by 18 x = 16833.33 So we have the 2nd truck with a load of 16833.33 kg, which is well over it's maximum permissible load of 8000 kg. Let's verify the results by plugging that mass into the before and after collision momentums. 60(5100 + 16833.33) = 60(21933.33) = 1316000 42(5100 + 4300 + 5100 + 16833.33) = 42(31333.33) = 1316000 They match. The 2nd truck was definitely over loaded.

6 0

3 years ago

It took a crew 9 h 36 min to row 8 km upstream and back again. If the rate of flow of the stream was 2 km/h, what was the rowing

babunello [35]

Answer:

3 km/h

Explanation:

Let's call the rowing speed in still water x, in km/h.

Rowing speed in upstream is: x - 2 km/h

Rowing speed in downstream is: x + 2 km/h

It took a crew 9 h 36 min ( = 9 3/5 = 48/5) to row 8 km upstream and back again. Therefore:

8/(x - 2) + 8/(x + 2) = 48/5 (notice that: time = distance/speed)