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melisa1 [442]
3 years ago
12

Calculate the average speed of a gazelle that runs 140 meters in 5.0 seconds

Physics
1 answer:
bekas [8.4K]3 years ago
3 0
140 m / 5 s = 28 meters per second 
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A circular loop of wire with 10 turns, radius 0.241 m and resistance 0.235 Ohms is connected to a 13.1 V power supply. The magne
SVETLANKA909090 [29]

Given Information:  

Resistance of circular loop = R = 0.235 Ω 

Radius of circular loop = r = 0.241 m

Number of turns = n = 10

Voltage = V = 13.1 V

Required Information:  

Magnetic field = B = ?  

Answer:  

Magnetic field = 0.00145 T

Explanation:  

In a circular loop of wire with n number of turns and radius r and carrying a current I induces a magnetic field B

B = μ₀nI/2r

Where μ₀= 4πx10⁻⁷ is the permeability of free space  and current in the loop is given by

I = V/R

I = 13.1/0.235

I = 55.74 A

B = 4πx10⁻⁷*10*55.74/2*0.241

B = 0.00145 T

Therefore, the magnetic field at the center of this circular loop is 0.00145 T

8 0
3 years ago
The amount of fluid displaced by a submerged object depends on its
Daniel [21]
The amount of fluid displaced by a submerged object depends on its volume.
7 0
3 years ago
Read 2 more answers
44. A rescue helicopter is hovering over a person whose boat has sunk. One of the rescuers throws a life preserver straight down
Vladimir [108]

Answer:

18.4 m

Explanation:

(a)

The known variables in this problem are:

u = 1.40 m/s is the initial vertical velocity (we take downward direction as positive direction)

t = 1.8 s is the duration of the fall

a = g = 9.8 m/s^2 is the acceleration due to gravity

(b)

The vertical distance covered by the life preserver is given by

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

If we substitute all the values listed in part (a), we find

d=(1.40 m/s)(1.8 s)+\frac{1}{2}(9.8 m/s^2)(1.8 s)^2=18.4m

8 0
3 years ago
Two teams of nine members each engage in a tug of war. Each of the first team’s members has an average mass of 68 kg and exerts
olga2289 [7]

Answer:

a)  a = - 0.106 m/s^2  (←)

b) T = 12215.1064 N

Explanation:

If

F₁ = 9*1350 N = 12150 N   (→)

F₂ = 9*1365 N = 12285 N  (←)

∑Fx = M*a = (M₁  +M₂)*a           (→)

F₁ - F₂ = (M₁  +M₂)*a        

→       a = (F₁ - F₂) / (M₁  +M₂ ) = (12150-12285)N/(9*68+9*73)Kg

→       a = - 0.106 m/s^2            (←)

(b) What is the tension in the section of rope between the teams?

If we apply  ∑Fx = M*a   for the team 1

F₁ - T = - M₁*a  ⇒   T = F₁ + M₁*a  

⇒   T = 12150 N + (9 * 68 Kg)*(0.106 m/s^2)

⇒ T = 12215.1064 N

If we choose the team 2 we get

- F₂ + T = - M₂*a  ⇒   T = F₂ - M₂*a  

⇒   T = 12285 N - (9 * 73 Kg)*(0.106 m/s^2)

⇒ T = 12215.1064 N

4 0
3 years ago
An 7.5 × binocular has 3.7-cm-focal-length eyepieces. What is the focal length of the objective lenses? Express your answer to t
elixir [45]

To solve this problem we will apply the concept of magnification, which is given as the relationship between the focal length of the eyepieces and the focal length of the objective. This relationship can be expressed mathematically as,

\mu = \frac{f_0}{f_e}

Here,

\mu = Magnification

f_e = Focal length eyepieces

f_0 = Focal length of the Objective

Rearranging to find the focal length of the objective

f_0 = \mu f_e

Replacing with our values

f_0 = 7.5* 3.7cm

f_0 = 27.75cm

Therefore the focal length of th eobjective lenses is 27.75cm

5 0
3 years ago
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