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ladessa [460]
2 years ago
15

Part A

Physics
1 answer:
geniusboy [140]2 years ago
4 0

Answer:

a) b = -5

b) slope = 3/2

Explanation:

a) The equation of a line is given as y = mx + b, where m is the slope of the line and b is the intercept on the y axis.

Given that y = 3x + b and it passes through the point (2, 1). Hence when x = 2, y = 1. Therefore, substituting for x and y:

1 = 3(2) + b

1 = 6 + b

b = 1 - 6

b = -5

b) The equation of a line passing through two points (x_1,y_1) and x_2,y_2 is given by:

y-y_1=\frac{y_2-y_1}{x_2-x_1}(x-x_1)

The equation of the line passing through the two points (0,3) and (4,9) is:

y-3=\frac{9-3}{4-0}(x-0)\\ \\y-3=\frac{3}{2}x\\ \\y = \frac{3}{2}x+3

Comparing y = (3/2)x + 3 with y = mx + b, the slope (m) is 3/2

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Billy picks up a 40 lb. dumbbell (mass = 18.14 kg). The center of his hand, where the dumbbell is held, is 56 cm (0.56 m) from t
umka21 [38]

Answer:

<h2>Force due to biceps is given as</h2><h2>F = 1991.05 N</h2>

Explanation:

For balancing the force we know that

Torque due to weight hold on his hand = torque due to force applied by biceps

So here we will have

mg \times L = F \times d

so we have

18.14 \times 9.8 \times 0.56 = F \times (0.05)

F = 1991.05 N

8 0
3 years ago
A 30 g bullet moving a horizontal velocity of 500 m/s comes to a stop 12 cm within a solid wall. (a) what is the change in the b
Sidana [21]
M = 30 g = 0.03 kg, the mass of the bullet
v = 500 m/s, the velocity of the bullet

By definition, the KE (kinetic energy) of the bullet is
KE = (1/2)*m*v²
      = 0.5*(0.03 kg)*(500 m/s)² = 3750 J
Because the bullet comes to rest, the change in mechanical energy is 3750 J.

The work done by the wall to stop the bullet in 12 cm is
W = (1/2)*(F N)*(0.12 m) = 0.06F J

If energy losses in the form of heat or sound waves are ignored, then
W = KE.
That is,
0.06F = 3750
F = 62500 N = 62.5 kN

Answer:
(a) 3750 J
(b) 62.5 kN

7 0
3 years ago
How much metabolic energy is required for a 68kg person to run at a speed of 15km/hr for 15min ?
irga5000 [103]
<span>What I have here is exactly the same problem, however, with the time changed to 19 mins:

metabolic energy = metabolic power*time = 1.150*19*60 = 1.311 kJ..corresponding to 1.311/4.186 = 313,2 Cal or kcal 

If we reasonably assume a metabolic eff.cy of 20%, it means we need to assume food for 1500 Cal approx.

Just plug the value t=15min to the equation and you will surely get the correct answer.

I hope my answer has come to your help. Thank you for posting your question here in Brainly. We hope to answer more of your questions and inquiries soon. Have a nice day ahead!
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8 0
3 years ago
A ball is thrown into the air with an upward velocity of 20 feet per second. Its height, h, in feet after t seconds is given by
Margarita [4]
Thank you for posting your question here at brainly. Feel free to ask more questions.  
<span>The best and most correct answer among the choices provided by the question is B. Reaches a max height of 8.25 feet after 0.63 seconds</span> .     <span><span>

</span><span>Hope my answer would be a great help for you. </span> </span>  

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3 0
3 years ago
The first car has twice the mass of a second car, but only half as much kinetic energy. When both cars increase their speed by 9
kirill [66]

Answer:

Speed of the car 1 =V_1=8.98m/s

Speed of the car 2 =V_2=17.96m/s

Explanation:

Given:

Mass of the car 1 , M₁ = Twice the mass of car 2(M₂)

mathematically,

M₁ = 2M₂

Kinetic Energy of the car 1 = Half the kinetic energy of the car 2

KE₁ = 0.5 KE₂

Now, the kinetic energy for a body is given as

KE =\frac{1}{2}mv^2

where,

m = mass of the body

v = velocity of the body

thus,

\frac{1}{2}M_1V_1^2=0.5\times \frac{1}{2}M_2V_2^2

or

\frac{1}{2}2M_2V_1^2=0.5\times \frac{1}{2}M_2V_2^2

or

2M_2V_1^2=0.5\times M_2V_2^2

or

2V_1^2=0.5\times V_2^2

or

4V_1^2= V_2^2

or

2V_1= V_2  .................(1)

also,

\frac{1}{2}M_1(V_1+9.0)^2=\frac{1}{2}M_2(V_2+9.0)^2

or

\frac{1}{2}2M_2(V_1+9.0)^2=\frac{1}{2}M_2(2V_1+9.0)^2

or

2(V_1+9.0)^2=(2V_1+9.0)^2

or

\sqrt{2}(V_1+9.0)=(2V_1+9.0)

or

(\sqrt{2}V_1+ \sqrt{2}\times 9.0)=(2V_1+9.0)

or

(\sqrt{2}V_1+ 12.72)=(2V_1+9.0)

or

(2V_1-\sqrt{2}V_1)=(12.72-9.0)

or

(0.404V_1)=(3.72)

or

V_1=8.98m/s

and, from equation (1)

V_2=2\times 8.98m/s = 17.96m/s

Hence,

Speed of car 1 =V_1=8.98m/s

Speed of car 2 =V_2=17.96m/s

8 0
2 years ago
Read 2 more answers
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