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Stels [109]
3 years ago
15

An object in static equilibrium has a coefficient of static friction as 0.110. If the normal force acting on the object is 95.0

newtons, what is the mass of the object?
Physics
1 answer:
igomit [66]3 years ago
7 0

Answer:

mass = 9.7 kg

Explanation:

As we know that when object is at rest on the ground of flat base then we will have

N - mg = 0

so from here

N = mg

now we have

N = 95 Newton

now from above equation we will have

95 = m\times g

95 = m\times 9.8

m = 9.7 kg

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A 75.0 kg person is on a Ferris Wheel that has a radius R = 16 m. The tangential velocity of the Ferris Wheel is 8.25 m/s. Calcu
Dovator [93]

Answer:

The period of rotation is

T=8.025s

Explanation:

The person is undergoing simple harmonic motion on the wheel

Given data

mass of the person =75kg

Radius of wheel r=16m

Velocity =8.25m/s

The oscillating period of simple harmonic motion is given as

T=(2*pi)/2=2*pi √r/g

Assuming that g=9.81m/s

Substituting our data into the expression we have

T=2*3.142 √ 16/9.81

T=6.284*1.277

T=8.025s

3 0
3 years ago
A football kicked in front of a goal post at an angle of 45 degree to the ground just clear the top by of the post 3m high. Calc
Firlakuza [10]

Answer:

A. 10.84 m/s

B. 1.56 s

Explanation:

From the question given above, the following data were obtained:

Angle of projection (θ) = 45°

Maximum height (H) = 3 m

Acceleration due to gravity (g) = 9.8 m/s²

Velocity of projection (u) =?

Time (T) taken to hit the ground again =?

A. Determination of the velocity of projection.

Angle of projection (θ) = 45°

Maximum height (H) = 3 m

Acceleration due to gravity (g) = 9.8 m/s²

Velocity of projection (u) =?

H = u²Sine²θ / 2g

3 = u²(Sine 45)² / 2 × 9.8

3 = u²(0.7071)² / 19.6

Cross multiply

3 × 19.6 = u²(0.7071)²

58.8 = u²(0.7071)²

Divide both side by (0.7071)²

u² = 58.8 / (0.7071)²

u² = 117.60

Take the square root of both side

u = √117.60

u = 10.84 m/s

Therefore, the velocity of projection is 10.84 m/s.

B. Determination of the time taken to hit the ground again.

Angle of projection (θ) = 45°

Velocity of projection (u) = 10.84 m/s

Time (T) taken to hit the ground again =?

T = 2uSine θ /g

T = 2 × 10.84 × Sine 45 / 9.8

T = 21.68 × 0.7071 / 9.8

T = 1.56 s

Therefore, the time taken to hit the ground again is 1.56 s.

7 0
3 years ago
What ia the law of convesation of energy​
Free_Kalibri [48]
In physics and chemistry, the law of conservation of energy states that the total energy of an isolated system remains constant; it is said to be conserved over time. This law means that energy can neither be created nor destroyed; rather, it can only be transformed or transferred from one form to another.
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The United States spends over $20 billion a year on space exploration through NASA. Do you think that this has been worth the co
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Sunday, July 20, marked 45 years since the United States put the first two astronauts safely on the moon. The cost for the Mercury, Gemini and Apollo programs was more than $25 billion at the time more like $110 billion in today’s world. The ensuing U.S. space efforts have cost an additional $196 billion for the shuttle and $50 billion for the space station. NASA’s total inflation-adjusted costs have been more than $900 billion since its creation in 1958 through 2014 (more than $16 billion per year). Looking back, have we gotten our money’s worth from the investment?

IamSugarBee

5 0
3 years ago
A huge tank of glycerine with a density of 1.260 g/cm3 is vertically stationed on a platform which is 15 m above the ground. The
EleoNora [17]

Answer:

The tank is losing 4.976*10^{-4}  m^3/s

v_g = 19.81 \ m/s

Explanation:

According to the Bernoulli’s equation:

P_1 + 1 \frac{1}{2} \rho v_1^2 + \rho gh_1 = P_2 +  \frac{1}{2}  \rho v_2^2 + \rho gh_2

We are being informed that both the tank and the hole is being exposed to air :

∴ P₁ = P₂

Also as the tank is voluminous ; we take the initial volume  v_1 ≅ 0 ;

then v_2 can be determined as:\sqrt{[2g (h_1- h_2)]

h₁ = 5 + 15 = 20 m;

h₂ = 15 m

v_2 = \sqrt{[2*9.81*(20 - 15)]

v_2 = \sqrt{[2*9.81*(5)]

v_2= 9.9 \ m/s  as it leaves the hole at the base.

radius r = d/2  = 4/2 = 2.0 mm

(a) From the law of continuity; its equation can be expressed as:

J = A_1v_2

J = πr²v_2    

J =\pi *(2*10^{-3})^{2}*9.9

J =1.244*10^{-4}  m^3/s

b)

How fast is the water from the hole moving just as it reaches the ground?

In order to determine that; we use the relation of the velocity from the equation of motion which says:

v² = u² + 2gh ₂

v² = 9.9² + 2×9.81×15

v² = 392.31

The velocity of how fast the water from the hole is moving just as it reaches the ground is : v_g = \sqrt{392.31}

v_g = 19.81 \ m/s

4 0
3 years ago
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