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Gre4nikov [31]
2 years ago
11

An object which has a mass of 70 kg is sitting on a cliff 10 m high. Calculate the object's Potential energy. Given g = 10m/s2

Physics
1 answer:
Rina8888 [55]2 years ago
6 0
Answer is a.)

Given: Mass m = 70 Kg   Height  h = 10 m;        g = 10 m/s²

Required. Gravitational Potential Energy or P.E

Formula: P.E = mgh

                     = (70 Kg)(10 m/s²)( 10 m)

              P.E = 7,000 Kg.m²/s² or

              P.E = 7,000 J

     
You might be interested in
A Porsche challenges a Honda to a 200-m race.Because the Porsche's acceleration of 3.5 m/s2 is larger than the Honda's 3.0 m/s2,
Blizzard [7]

Answer:

Honda won by 0.14 s

Explanation:

We are given that

Distance =S=200 m

Initial velocity of Honda=u=0m/s

Initial velocity of Porsche=u'=0m/s

Acceleration of Honda=3.0m/s^2

Acceleration of Porsche's=3.5m/s^2

Time taken by Honda  to start=1 s

s=ut+\frac{1}{2}at^2

Substitute the values

200=0(t)+\frac{1}{2}(3)t^2

200=\frac{3}{2}t^2

t^2=\frac{200\times 2}{3}=\frac{400}{3}

t=\sqrt{\frac{400}{3}}=11.55s

Time taken by Honda=11.55 s

Now, time taken by  Porsche

200=\frac{1}{2}(3.5)t^2

t^2=\frac{200\times 2}{3.5}

t=\sqrt{\frac{400}{3.5}}=10.69 s

Total time taken by Porsche=10.69+1=11.69 s

Because it start 1 s late

Time taken by Honda is less than Porsche .Therefore, Honda won and

Time =11.69-11.55=0.14 s

Honda won by 0.14 s

3 0
2 years ago
Determine the force P required to maintain the 200-kg engine in the position for which θ = 30°. The diameter of the pulley at B
gregori [183]

Answer:

The force P required  is 1759.22 N

Explanation:

The missing diagram is seen in the first image below.

From the second image, we can see the schematic diagram of the engine hanging over the pulley.

To start with determining the value of the angle ∝;

tan \ \alpha = \dfrac{CD}{BD}

where;

BD = AB-AD

Then;

tan \ \alpha = \dfrac{CD}{AB-AD}

\alpha = tan^{-1} \bigg(\dfrac{CD}{AB-AD} \bigg )

replacing their respective values, where;

CD = 2 sin 30° m,  AB = 2m and AD = 2 cos 30° m

\alpha = tan^{-1} \bigg(\dfrac{2 \ sin \ 30^0}{2-2 \ cos \ 30^0} \bigg )

\alpha = tan^{-1} \bigg(\dfrac{1}{2-1.732} \bigg )

\alpha = tan^{-1} \bigg(\dfrac{1}{0.268} \bigg )

\alpha = tan^{-1} \bigg(3.73\bigg )

\alpha \simeq 75^0

From the third diagram attached below:

The tension occurring in the thread BC is equal to force P

T_{BC} = P

Using the force equilibrium expression along the horizontal direction.

\sum F_x = 0\\\\ -T_{AC} \ cos \ 30^0 + Pcos \alpha = 0

replacing the value of \alpha \simeq 75^0

-T_{AC} \  cos 30^0 + P cos 75^0  = 0

P \ cos \ 75^0 = T_{AC} \ cos \ 30^0

P  =\dfrac{ T_{AC} \ cos \ 30^0}{\ cos \ 75^0} \ \ \ - - -  (1)

Along the vertical direction, the force equilibrium equation can be expressed as:

\sum F_y =0

-W + P \ sin \alpha + T_{AC} \ sin \ 30^0  = 0

W = P \ sin \ \alpha + T_{AC} \ sin \ 30^0

replacing \alpha \simeq 75^0 and P  =\dfrac{ T_{AC} \ cos \ 30^0}{\ cos \ 75^0}

W =\dfrac{T_{AC} \ cos \ 30^0}{cos \ 75^0}\times sin \ 75^0 + T_{AC} \ sin \ 30^0

Also, replacing W for (200 × 9.81) N

200 \times 9.81 =\dfrac{T_{AC} \ cos \ 30^0}{cos \ 75^0}\times sin \ 75^0 + T_{AC} \ sin \ 30^0

200 \times 9.81 = T_{AC} \ cos \ 30^0 \ tan \ 75^0 + T_{AC} \ sin \ 30^0

1962= T_{AC} \ ( cos \ 30^0 \ tan \ 75^0 + \ sin \ 30^0)

1962= T_{AC} \ (0.8660\times 3.732 + 0.5)

1962= T_{AC} \ (3.231912 + 0.5)

1962= T_{AC} \ (3.731912)

T_{AC}  = \dfrac{1962}{ \ (3.731912)}

T_{AC}  = 525.736 \ N

From P  =\dfrac{ T_{AC} \ cos \ 30^0}{\ cos \ 75^0}

P =\dfrac{ 525.736 \ cos \ 30^0}{\ cos \ 75^0}

P =\dfrac{ 525.736 \times0.866}{0.2588}

P = 1759.22 N

Thus, the force P required  is 1759.22 N

6 0
1 year ago
Air has a density of 1.3 kg/m³. Calculate the mass of 36 m³ of air in kilograms. Give your answer to 1 decimal place.
vredina [299]

Answer:

46.8 kg

Explanation:

Mass = (density)(volume)

= (1.3)(36)

<u>M</u><u>a</u><u>s</u><u>s</u><u> </u><u>=</u><u> </u><u>4</u><u>6</u><u>.</u><u>8</u><u> </u><u>k</u><u>g</u>

3 0
2 years ago
How would the interference pattern change for this experiment if a. the grating was moved twice as far from the screen and b. th
Airida [17]

Answer:

See explanation

Explanation:

Solution:-

- Here we will assume that the grating has the line density ( N ) defined by the number of lines per mm.

- The angle that each fringe forms on the screen is defined by ( θ ).

- The order of bright/dark spot is defined by an integer ( n )

- The wavelength of the incident light is ( λ )

- Here we will use the relation given for diffraction grating by the Young's Experiment as follows:

                               n*lambda = \frac{sin(theta)}{N}

- The above given formulation is for constructive interference.

- We will inspect the effect of increasing the distance between the screen and the grating. Consider the length ( L ) from the center of the grating to the center of the screen. The distance ( yn ) will denote the distance between each fringe in vertical direction on the screen.

- For small angles ( θ ) we can make an approximation of sin ( θ ) ≈ tan ( θ ). Where,

                            sin ( θ ) ≈ tan ( θ ) = [ yn / L ]

- Substitute the above approximation in the given relation of diffraction gratings as follows:

                            y_n = n*lamda*L*N

- To double the distance between the screen and grating we will use the above relation with ( 2L ):

                            yn ∝ L

Result: The distance between each order of bright and dark fringe is doubled. The interference pattern would have twice the spread! This also means that less number of bright spots would be seen on the screen as the coverage area would require a larger screen to accommodate the entire interference pattern. The spread also reduces the intensity/contrast between the bright and dark fringes because the distance travelled by each ray of light has increased. The intensity is inversely proportional to the square of distance travelled.

- Similarly, the line density of the grating ( N ) was doubled. Then,

                            yn ∝ N

Result: The distance between each order of bright and dark fringe is doubled. The interference pattern would have twice the spread!This also means that less number of bright spots would be seen on the screen as the coverage area would require a larger screen to accommodate the entire interference pattern.

4 0
2 years ago
You have two square metal plates with side lengths of (6.50 C) cm. You want to make a parallel-plate capacitor that will hold a
gtnhenbr [62]

Answer:

The necessary separation between  the two parallel plates is 0.104 mm

Explanation:

Given;

length of each side of the square plate, L = 6.5 cm = 0.065 m

charge on each plate, Q = 12.5 nC

potential difference across the plates, V = 34.8 V

Potential difference across parallel plates is given as;

V = \frac{Qd}{L^2 \epsilon_o} \\\\d = \frac{V L^2 \epsilon_o}{Q}

Where;

d is the separation or distance between the two parallel plates;

d = \frac{VL^2 \epsilon_o}{Q} \\\\d =  \frac{34.8*(0.065)^2 *8.854*10^{-12}}{12.5*10^{-9}} \\\\d = 0.000104 \ m\\\\d = 0.104 \ mm

Therefore, the necessary separation between  the two parallel plates is 0.104 mm

6 0
2 years ago
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