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Katarina [22]
1 year ago
12

( Find the value of the following in ms-1 54kmhr

Physics
1 answer:
Darina [25.2K]1 year ago
6 0

Answer:

54 × 5/18 = 15m/s

Explanation:

to convert km/hr to m/s you multiply by 5/18

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The Biot-Savart Law describes the relationship between which of the following? Select all that apply.
faltersainse [42]
The correct answer is magnetic field, electric field, and charges.


8 0
3 years ago
A pen rolls off a 0.55–meter high table with an initial horizontal velocity of 1.2 meters/second. At what horizontal distance fr
NARA [144]
To find the horizontal distance multiple the horizontal velocity by the time. Since there is no given time it must be calculated using kinematic equation.

Y=Yo+Voyt+1/2at^2
0=.55+0+1/2(-9.8)t^2
-.55=-4.9t^2
sqrt(.55/4.9)=t
t=0.335 seconds

Horizontal distance
=0.335s*1.2m/s
=0.402 meters
8 0
3 years ago
A force F~ = Fx ˆı + Fy ˆ acts on a particle that
Hatshy [7]

Answer:

W = 46 J

Explanation:

We need to find the angle between the two vectors Force vector and displacement vector.

First we will find the angle α of the force vector

tan\alpha =\frac{1}{8} \\\\\\alpha =7.125 deg\\

Then we find the angle β of the displacement vector

tan\beta=\frac{2}{6} \\\\beta = 18.43 deg\\

With these two angles we can find the angle between the two vectors

∅ = α + β = 25.56 deg

The definition of work is given by the expression

W=F*d*cos (theta)

The absolute value of F will be:

F=\sqrt{8^{2}+1^{2}  } \\F= 8.06 N

The absolute value of d will be:

d=\sqrt{(6 )^{2}+(2)^{2}  } \\d= 6.32m\\

Now we have:

W=8.06*6.32*cos(25.56)\\W=46 J

4 0
3 years ago
calculate the work done in kilo joules in lifting a mass of 20kg at steady velocity through a vertical height of 20m
bulgar [2K]

Answer:

\huge\boxed{\sf Work\ done = 4 kJ}

Explanation:

Since work done is in the form of potential energy, we will use the formula of potential energy here.

We know that,

<h3>P.E. = mgh </h3>

Where,

m = mass = 20 kg

g = acceleration due to gravity = 10 m/s²

h = vertical height = 20 m

So,

<h3>Work done = mgh</h3>

Work done = (20)(10)(20)

Work done = 4000 joules

Work done = 4 kJ

\rule[225]{225}{2}

5 0
2 years ago
HELP!!!!!
devlian [24]

Answer:

c

Explanation:

6 0
3 years ago
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