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Aloiza [94]
3 years ago
13

interval (from lowering the body to his feet taking off) lasts for 0.3 s and the mass of the player is 90 kg. Ignore air resista

nce. Part A Determine the force that the the player exerts on the surface when he jumps. Express your answer to one significant figure and include the appropriate units. Enter positive value if the force is upward and negative value if the force is downward.
Physics
1 answer:
steposvetlana [31]3 years ago
8 0

Answer:

The force applied to the surface is 9 kilo Newton.

Explanation:

While jumping on the surface the player applies the force that is equal to its weight on the surface.

The mass of the player is given as 90 kg.

Force applied by the player = weight of the player

Force applied by the player = m × g

Where m is the mass of the player and g is acceleration due to gravity

Force applied by the player = 90 × 9.8

Force applied by the player = 882 Newton

Expressing your answer to one significant figure, we get

Force applied by the player =0. 9 kilo Newton

The force applied to the surface is 0.9 kilo Newton.

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inna [77]

Answer:

Student C

Explanation:

order from heaviest to lightest is...

9 kg (A) , 8000g (8 kg) (D) , 2800g (2.8kg) (B), 2 kg (C)

5 0
3 years ago
An electric dipole is formed from ± 5.0 nC point charges spaced 3.0 mm apart. The dipole is centered at the origin, oriented alo
Ymorist [56]

Answer:

The electric field strength at point (x,y) = ( 20 mm ,0cm) is =<u>16321.0769 N/C</u>

The electric field strength at point (x,y) = (0cm, 20 mm) is =<u>35321.58999 N/C</u>

Explanation:

Question: What is the electric field strength at point (x,y) = ( 20 mm ,0cm)?

Answer:

The electric field at any given point of the dipole is given as:

E= (KP) ÷ (r^2 + a^2)^3/2

Where:

K = 9x10^9 Nm^2/c^2 (coloumb constant)

P = (0.003) (5x10^-9c) which is the movement of the dipole

(0.003) is arrived at when mm is converted to m. 3.0 mm space apart was converted to a meter.

r= the point, in the question above is 20mm = 0.02m

Now, the electric field, E can be calculated by putting the values in the formula above:

E = (KP) ÷ (r^2 + a^2)^3/2

= (9x10^9 Nm^2/c^2) (0.003 m) (5x10^-9c) ÷ [ (0.02m)^2 + (0.003)^2]^3/2

= 0.135 ÷ (8.271513x10^-6)

=<u>16321.0769 N/C</u>

 Question: What is the electric field strength at point (x,y) = (0cm, 20 mm )?

Answer:

Here, the electric field, E= 2krp ÷ (r^2 - a^2)^2

E= 2 (9x10^9 Nm^2/c^2) (0.02m) (0.003 m) (5x10^-9c) ÷ [(0.02m)^2 - (0.003)^2]^2

= 0.0054 ÷  0.000000152881

=<u>35321.58999 N/C</u>

8 0
2 years ago
Which of the following statements apply to electric charges?
Gre4nikov [31]

Answer:

The statement "If a positively charged rod is brought close to a positively charged object, the two objects will repel " applies to electric charges.

Explanation:

There are only two types of electric charges. Both having own magnitude but different charge.

1. Positive charge

2. Negative charge

Like charges repel each other and opposite charges always attract each other.

When a positively charged rod is brought close to a positively charged object, the rod and the object will repel.

6 0
3 years ago
How do we represent Earth’s face?
jeka94

Answer:

Human-driven changes in arrive utilize and arrive cover such as deforestation, urbanization, and shifts in vegetation designs moreover change the climate, coming about in changes to the reflectivity of the Soil surface (albedo), emanations from burning timberlands, urban warm island impacts and changes within the normal water cycle.

6 0
3 years ago
A transmission diffraction grating with 420 lines/mm is used to study the light intensity of di event orders (n). A screen is lo
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Answer:

Explanation:

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= 2.38 x 10⁻³ mm

2.38 x 10⁻⁶ m

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= 680 x 10⁻⁹ m

For bright red band

position x on the screen

= n λD / a ,         n = 0,1,2,3 etc

D = distance of screen

putting n = 1 , 2 and 3 , we can get three locations of bright red band.

x₁ = λD / a

=  680 x 10⁻⁹ x 2.8 / 2.38 x 10⁻⁶

= .8 m

= 80 cm

Position of second bright band

= 2 λD / a

= 2 x 80

= 160 cm

Position of third bright band

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= 240 cm

5 0
3 years ago
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