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Novosadov [1.4K]
3 years ago
14

Explain, using your own example why you must always give a unit when reporting a measurement.

Physics
1 answer:
Ludmilka [50]3 years ago
5 0
If you don't give unit with your magnitude (number) then, anyone can't figure it what you are talking about!
Suppose, you said, the weight of an iron bar is 10.
What does that mean? We can't even figure out how heavier is it. 

Now, If you tell that the weight of a person is 10 Newton, then it makes sense 'cause 10 N has specific value. When you wrote 10 then, it could be newton, dyne or any other derived unit.

Hope this helps!
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Can someone help me answer this??
Bingel [31]

Answer:

<h2>3,2 oky na dekh Lena ek bar</h2><h2>2,5</h2>
7 0
3 years ago
Two blocks, joined by a string, have masses of 6.0 kg and 9.0 kg. They rest on a frictionless horizontal surface. A 2nd string,
Gennadij [26K]

Answer:

Explanation:

30 N force is pulling total mass of 15 kg , so acceleration in the system of masses

= 30 / 15

= 2 m / s²

Let us now consider forces acting on 9 kg . 30 N is pulling it in forward direction . Tension T in the string attached to it is pulling it in reverse direction

so net force on it

30 - T

Applying Newton's law of motion on it

30 - T = mass x acceleration

30 - T = 9  x 2

30 - 18 = T

T = 12 N

4 0
4 years ago
Interactive LearningWare 4.1 reviews the approach taken in problems such as this one. A 1800-kg car is traveling with a speed of
Lubov Fominskaja [6]

Answer:

F= 4788 N

Explanation:

Because the car moves with uniformly accelerated movement we apply the following formula:

vf²=v₀²+2*a*d Formula (1)

Where:  

d:displacement in meters (m)  

v₀: initial speed in m/s    

vf: final speed in m/s  

a: acceleration in m/s²

Data

d=36.9 m

v₀=14.0 m/s m/s    

vf= 0  

Calculating of the acceleration of the car

We replace dta in the formula (1)

vf²=v₀²+2*a*d

(0)²=(14)²+2*a*(36.9)

-(14)²= (73.8) *a

a= - (196) /  (73.8)

a= - 2.66 m/s²

Newton's second law of the car in direction  horizontal (x):

∑Fx = m*ax Formula (2)

∑F : algebraic sum of the forces in direction x-axis (N)

m : mass (kg)

a : acceleration  (m/s²)

Data

m=1800 Fkg

a= - 2.66 m/s²

Magnitude of the horizontal net force (F) that is required to bring the car to a halt in a distance of 36.9 m :

We replace data in the formula (2)

-F= (1800 kg) * ( -2.66 m/s² )

F= 4788 N

6 0
3 years ago
A conductor shaped as a circular loop with a radius of 4.0 m is located in a uniform but changing magnetic field. If the maximum
bekas [8.4K]

Answer:

\frac{\delta B}{\delta t}= 0.0995 \  T/s

Explanation:

Given that :

The radius of the circular loop = 4.0 m

Maximum Emf E_{max} = 5.0 V

The  maximum rate at which the magnetic field strength is changing if the magnetic field is oriented perpendicular to the plane in which the loop lies can be determined via the expression;

E_{max} = Area (A) * \frac{\delta B}{\delta t}

E_{max} = \pi r^2 * \frac{\delta B}{\delta t}

5.0 = \pi * (4.0)^2 * \frac{\delta B}{\delta t}

5.0 = 50.27 * \frac{\delta B}{\delta t}

\frac{\delta B}{\delta t}= \frac{5}{50.27}

\frac{\delta B}{\delta t}= 0.0995 \  T/s

5 0
3 years ago
Consult Interactive Solution 7.10 for a review of problem-solving skills that are involved in this problem. A stream of water st
AlekseyPX

Explanation:

Initial speed of the incident water stream, u = 16 m/s

Final speed of the exiting water stream, v = -16 m/s

The mass of water per second that strikes the blade is 48.0 kg/s.

We need to find the magnitude of the average force exerted on the water by the blade. The force acting on an object is given by :

F=\dfrac{m(v-u)}{t}

Here, \dfrac{m}{t}=48\ kg/s

F=48\times (-16-16)\ N\\\\F=-1536\ N\\\\|F|=1536\ N

So, the magnitude of the average force exerted on the water by the blade is 1536 N.

5 0
3 years ago
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