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Assoli18 [71]
3 years ago
7

To convert minutes per second into kilometre per hour we multiply the speed with​

Physics
1 answer:
ankoles [38]3 years ago
3 0

Answer:

To convert m/sec into km/hr, multiply the number by 18 and then divide it by 5.

Explanation:

please mark as brainliest

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A box is 30 cm wide,40cm long and 25cm high calculate:
Leviafan [203]

Answer:

1200 cm^2

Explanation:

Hope it is helpful.....

3 0
3 years ago
What activities areaways to overcome barriers to physical fitness
dexar [7]

Answer:

No Time

Schedule activities into your day and use an exercise log so you can see how little time it takes.

Build activities into everyday tasks no matter where you are: Bike to work. Use the stairs. Take walk breaks at work. Garden. ...

Find an activity you enjoy that works for your schedule

4 0
3 years ago
A box of mass 20kg is pulled up an inclined plane by a force of 285N. Given that the value of
lorasvet [3.4K]

Answer:

19.6 m/s

Explanation:

The parameters given are:

Mass M = 20 Kg

Force F = 285 N

Angle Ø = 30 degree

Time t = 4 seconds

Coefficient of friction = 0.72

At the plane, the weight of the box will be mgsinØ

Resolving forces at the plane, we will have:

MgsinØ + Fr = F

Where Fr = frictional force.

Fr = F - mgsinØ

Substitute all the parameters into the formula

Fr = 285 - 20 × 9.8 sin30

Fr = 285 - 98

Fr = 187 N

But for the box moving toward the top of the plane,

F - Fr = ma

Where a = V/t

Substitute all the parameters involved into the formula

285 - 187 = 20 ( V/4)

98 = 5V

V = 98/5

V = 19.6 m/s

Therefore, the speed with which the box is moving is 19.6 m/s

4 0
3 years ago
What is the resultant displacement of the two vectors below?
andriy [413]

Answer:

C 3.6 cm, 56 degrees North of the East axis

Explanation:

The two vectors are perpendicular to each other, so we can find the magnitude of their resultant simply by using the Pythagorean theorem:

R=\sqrt{A^2+B^2}

where

A = 2.0 cm is the magnitude of the first vector

B = 3.0 cm is the magnitude of the second vector

Substituting,

R=\sqrt{2^2+3^2}=3.6 cm

Now we have to find the angle. If we measure the angle as North of East, the tangent of the angle is equal to the ratio between the component along North and the component along East. Therefore, in this case:

tan \theta = \frac{B}{A}=\frac{3}{2}=1.5\\\theta = tan^{-1}(1.5)=56.3 \sim 56^{\circ}

So, 56 degrees North of East.

4 0
4 years ago
A centrifuge has an angular velocity of 3,000 rpm, what is the acceleration (in unit of the earth gravity) at a point with a rad
Anna71 [15]

Answer:

a_{r} = 1006.382g \,\frac{m}{s^{2}}

Explanation:

Let suppose that centrifuge is rotating at constant angular speed, which means that resultant acceleration is equal to radial acceleration at given radius, whose formula is:

a_{r} = \omega^{2}\cdot R

Where:

\omega - Angular speed, measured in radians per second.

R - Radius of rotation, measured in meters.

The angular speed is first determined:

\omega = \frac{\pi}{30}\cdot \dot n

Where \dot n is the angular speed, measured in revolutions per minute.

If \dot n = 3000\,rpm, the angular speed measured in radians per second is:

\omega = \frac{\pi}{30}\cdot (3000\,rpm)

\omega \approx 314.159\,\frac{rad}{s}

Now, if \omega = 314.159\,\frac{rad}{s} and R = 0.1\,m, the resultant acceleration is then:

a_{r} = \left(314.159\,\frac{rad}{s} \right)^{2}\cdot (0.1\,m)

a_{r} = 9869.588\,\frac{m}{s^{2}}

If gravitational acceleration is equal to 9.807 meters per square second, then the radial acceleration is equivalent to 1006.382 times the gravitational acceleration. That is:

a_{r} = 1006.382g \,\frac{m}{s^{2}}

6 0
3 years ago
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