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Alexxandr [17]
3 years ago
6

Match the following. Column A 1. Torque 2. Centre of gravity 3. Plumb line Column B A. Line of centre of gravity B. Maximum cons

tant velocity C. Not in motion D. Point of action of weight E. Turning effect of force F. Study of forces 4. Terminal velocity 5. Statics​
Physics
1 answer:
gregori [183]3 years ago
8 0

Answer:

1. Torque → F. Study of forces

2. C.O.G → D. Point of action of weight.

3. Plumb line → A. Line of C.O.G

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Gold has a density of 19300 kg/m³. Calculate the mass of 0.02 m³ of gold in kilograms.
hjlf

Answer:

Mass = 386 kg

Explanation:

<u><em>Density = Mass / Volume</em></u>

Mass = Density × Volume

Where D = 19300 kg/m³ , V = 0.02 m³

<em>Putting the given in the above formula</em>

Mass = 19300 × 0.02

Mass = 386 kg

7 0
3 years ago
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According to the picture, which is the least dense?
alexdok [17]

Answer:

a the chess peice

Explanation:

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A spring with constant k = 78 N/m is at the base of a frictionless, 30.0°-inclined plane. A 0.50-kg block is pressed against the
olasank [31]

Explanation:

The given data is as follows.

   Spring constant (k) = 78 N/m,     \theta = 30^{o}

 Mass of block (m) = 0.50 kg

According to the formula of energy conservation,

                mgh sin \theta = \frac{1}{2}kx^{2}

       h = \frac{1}{2} \times \frac{kx^{2}}{mg Sin \theta}

          = \frac{78 \times 0.04}{2 \times 0.5 \times 9.8 \times 0.5}

          = 0.64 m

Thus, we can conclude that the distance traveled by the block is 0.64 m.

4 0
3 years ago
A car moves with constant acceieration of -0.s m/s? on a straight portion of the road. Att- 0s the car has a velocity of 69 mph,
Crazy boy [7]

Answer:

b) d = 0.71 Km

Explanation:

Car kinematics

Car 1 moves with uniformly accelerated movement

v_f^2=v_o^2+2a*d Formula (1)

d: displacement in meters (m)

v₀: initial speed in m/s

vf: final speed in m/s

a: acceleration in m/s²

Equivalences:

1mile = 1609.34 meters

1 hour = 3600s

1km = 1000m

Known data

v_o = 69\frac{mile}{hour} *1609.34\frac{meter}{mile} *\frac{1}{3600}\frac{hour}{s}=30.8 \frac{m}{s}

v_f= \frac{69}{2} \frac{mile}{hour} =34.5\frac{mile}{hour}=15.4 \frac{m}{s}

a = -0.5 m/s²

Distance calculation

We replace data in the Formula (1)

15.4^2 = 30.8^2+2(-0.5)*d

2(0.5)*d = 30.8^2 - 15.4^2

d =\frac{ 30.8^2 - 15.4^2}{2(0.5) }= 717.6m

d = 717.6 m\frac{1km}{1000m} = 0.7176Km

8 0
3 years ago
A typical cell has a membrane potential of -70 , meaning that the potential inside the cell is 70 less than the potential outsid
Oxana [17]

Answer:

energy is stored is 2.2 × 10⁻¹³ J

Explanation:

The capacitance  of the cell is given with the expression

C = (KE₀A) / d

k is the dielectric constant, A is the area of the cell, d is the thickness of the cell.

Now given that; the diameter is 50,

Area A = 4πR²

A = 4π × ( 25 × 10⁻⁶ m)²

A = 7850×10⁻¹² m²

our capacitance C = (KE₀A) / d

C = [9 ( 8.85 × 10⁻¹² C²/N.m²  × 7850×10⁻¹² m² )] / 7×10⁻⁹ m

C = 8.93 × 10⁻¹¹ F

Now Energy stored

E = 1/2 × CV²

E = 1/2 × (8.93 × 10⁻¹¹ F) × ( 70 × 10⁻³ V)²

E = 2.2 × 10⁻¹³ J

Therefore energy is stored is 2.2 × 10⁻¹³ J

8 0
3 years ago
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