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pochemuha
4 years ago
8

Match the measurement with the proper Sl unit Acceleration: Velocity Distance:

Physics
2 answers:
Ket [755]4 years ago
6 0

Answer:

Acceleration - ms^-2

Velocity- ms^-1

Distance-m

Xelga [282]4 years ago
3 0

Answer:

Acceleration:

m/s^{2}

Velocity:

m/s

Distance:

m

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Can we use the traditional methods of measurement? Are they reliable? If not why?​
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Answer:

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4 0
3 years ago
What are the answers for numbers 2 & 3?
ivolga24 [154]
The answer is 6 i no this because im in high school this app is for high schoolers now I'll answer more. of ur questions if u make them have more points now u shouldn't play this game because ur prob in elementary but any ways the answer is 6
8 0
3 years ago
What is the minimum force required to increase the energy of a car by 84 J over a distance of 38 m? Assume the force is constant
konstantin123 [22]

Answer:

2.210N

Explanation:

Workdone = Force x distance

Distance = 38m , Workdone = 84J

Hence 84J = Force x 38m

Force = 84J / 38m

Force = 2.210N =2.2N

4 0
3 years ago
A solid cylinder is released from the top of an inclined plane of height 0.50 m. From what height, in meters, on the incline sho
dangina [55]

Answer:

Explanation:

for rolling motion down the plane acceleration is given by the following expression

a = g sinθ / (1 + k² / R²)

here k is radius of gyration and R is radius of the object rolling down .

for cylinder I = 1/2 m R²

so k² = R² / 2

k² / R² = 1/2

a = g sinθ /( 1 + 1 / 2 )

= 2 / 3 x  g sinθ

v = √ 2 a s

= √ (2 x  2 / 3 x  g sinθ s )

= √ (4  / 3 x  g h  )

= √ (4  / 3 x  g x .5  )

= √ 2g / 3

for sphere  I = 2/5  m R²

so k² = 2/5 R²

k² / R² = 2 / 5  

a = g sinθ / (1 + 2 / 5)  

= 5 / 7  x  g sinθ

v = √ 2 a s

= √ (2 x  5 / 7  x  g sinθ s )

= √ (10/7  x  g h  )

Given

√ (10/7  x  g h  ) = √ 2g / 3

10/7  x  g h  = 2g / 3

h = 14 / 30 m

= .47 m .

5 0
3 years ago
A body of mass 400 kg is suspended at a lower end of a light vertical chain and is being pulled up vertically. Initially the bod
yKpoI14uk [10]

Answer:31.62 m/s

Explanation:

Given

mass of body m=400 kg

Pull on chain is F_1=6000g N=60 kN

Pull get smaller at the rate of F_2=360g N/m

Net Upward Force F=6000 g-360 g\times 10=24 kN

net acceleration a=\frac{F}{m}

a=\frac{24\times 1000}{m}

a=\frac{24000}{400}

a=60 m/s^2

but g is acting downward

a_{net}=a-g=60-10=50 m/s^2

using v^2-u^2=2 as

here initial velocity is zero

v^2=2\times 50\times 10

v=31.62 m/s

7 0
4 years ago
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