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kobusy [5.1K]
3 years ago
5

How can you prove that acceleration is a derived unit​

Physics
2 answers:
saul85 [17]3 years ago
8 0

Answer:

Explanation:

Acceleration is derived unit because it has two fundamental units involved i.e. meter and second square.

Aliun [14]3 years ago
7 0

a = (dx / dt)²

Explanation:  Unit of distance is m (metres) and unit of time is s (seconds) speed v is  first derivative of distance x versus time:

  v = dx / dt, unit is m/s.    Acceleration is second derivative of

 speed versus time  a = (dx / dt)² = (dv/dt) , unit is m/s²

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Rasek [7]
<span>the arrangement of the outer planets is 
</span>1. Mercury 
<span>2. Venus </span>
<span>3. Earth </span>
<span>4. Mars </span>
<span>5. Jupiter </span>
<span>6. Saturn </span>
<span>7. Uranus </span>
8. Neptune
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4 0
3 years ago
Returning once again to our table top example of a horizontal mass on a low-friction surface with m = 0.254 kg and k = 10.0 N/m
Julli [10]

Explanation:

Given that,

Mass = 0.254 kg

Spring constant [tex[\omega_{0}= 10.0\ N/m[/tex]

Force = 0.5 N

y = 0.628

We need to calculate the A and d

Using formula of A and d

A=\dfrac{\dfrac{F_{0}}{m}}{\sqrt{(\omega_{0}^2-\omega^{2})^2+y^2\omega^2}}.....(I)

tan d=\dfrac{y\omega}{(\omega^2-\omega^2)}....(II)

Put the value of \omega=0.628\ rad/s in equation (I) and (II)

A=\dfrac{\dfrac{0.5}{0.254}}{\sqrt{(10.0^2-0.628)^2+0.628^2\times0.628^2}}

A=0.0198

From equation (II)

tan d=\dfrac{0.628\times0.628}{((10.0^2-0.628)^2)}

d=0.0023

Put the value of \omega=3.14\ rad/s in equation (I) and (II)

A=\dfrac{\dfrac{0.5}{0.254}}{\sqrt{(10.0^2-3.14)^2+0.628^2\times3.14^2}}

A=0.0203

From equation (II)

tan d=\dfrac{0.628\times3.14}{((10.0^2-3.14)^2)}

d=0.0120

Put the value of \omega=6.28\ rad/s in equation (I) and (II)

A=\dfrac{\dfrac{0.5}{0.254}}{\sqrt{(10.0^2-6.28)^2+0.628^2\times6.28^2}}

A=0.0209

From equation (II)

tan d=\dfrac{0.628\times6.28}{((10.0^2-6.28)^2)}

d=0.0257

Put the value of \omega=9.42\ rad/s in equation (I) and (II)

A=\dfrac{\dfrac{0.5}{0.254}}{\sqrt{(10.0^2-9.42)^2+0.628^2\times9.42^2}}

A=0.0217

From equation (II)

tan d=\dfrac{0.628\times9.42}{((10.0^2-9.42)^2)}

d=0.0413

Hence, This is the required solution.

5 0
3 years ago
Object A attracts object B with a gravitational force of 5 newtons from a given distance. If the distance between the two object
enyata [817]

Answer:

10 newtons, because the gravitaional force willbe stronger the closer it gets.

7 0
3 years ago
Identify the object that has the largest moment from the list of objects given below.
Luda [366]

Answer:

A fired bullet

Explanation:

A fired bullet is faster than the speed of sound

7 0
3 years ago
A crane is used to lower weights into a lake for an underwater construction project. Determine the tension in the cable of the c
atroni [7]

Answer:

Tension in the cable = 591292.8 N or 591.3 kN

Explanation:

Tension in the cable = Weight of object in air - upthrust/bouyancy due to the liquid

Weight of object in air = mass * acceleration due to gravity, g

Since mass is not given, it is obtained from the formula, mass = density * volume

Volume of the sphere = 4/3πr³ where r, radius = 2 m; π = 22/7

volume = 4/3 * 22/7 * (2m)³

volume of sphere = 33.52 m³

Density of concrete sphere = specific gravity * density of water

where, specific gravity = 2.8, density of water = 1000 kg/m³

density of concrete =  2.8 * 1000 kg/m³

density of concrete = 2800 kg/m³

acceleration due to gravity, g = 9.8 m/s²

Thus, weight of concrete = 2800 kg/m³ * 33.52 m³ * 9.8 m/s² = 919788.8 N

Upthrust = density of water * volume of sphere * g

Upthrust = 1000 kg/m³ * 33.52 m³ * 9.8 m/s² = 328496 N

The tension in the cable is the calculated as below;

Tension in the cable = Weight of object in air - upthrust/bouyancy due to the liquid

Tension in cable = 919788.8 N - 328.496 N = 591292.8 N

Therefore, tension in the cable is 591292.8 N or 591.3 kN

5 0
3 years ago
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