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3 years ago

9

Anthony and Maelynn are watching a football game outside on a sunny day. Anthony is wearing a black shirt and Maelynn is wearing

a white shirt. Which person will be warmer after the game? Explain your answer.

2 answers:

melomori [17]3 years ago

7 0

Answer: Anthony will be warmer after the game.

Explanation :

Anthony and Maelynn are watching a football game outside on a sunny day. Anthony is wearing a black shirt and Maelynn is wearing a white shirt. Anthony will be warmer after the game. The black color is a good absorber of radiation and a bad reflector.

The black color absorbs heat until a thermal equilibrium is attained. So, it is advisable to wear cotton clothes in summers not dark colored clothes.

astra-53 [7]3 years ago

3 0

Answer:

Anthony will be warmer after the game because he is wearing a black shirt. Dark colors absorb more thermal energy than light colors.

Explanation:

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Ivan

yes you are correct about atoms

7 0

3 years ago

Suppose that the position of a particle is given by s=f(t)=5t3+6t+9. (a) Find the velocity at time t.

atroni [7]

This question is incomplete, the complete question is;

Suppose that the position of a particle is given by s=f(t)=5t³ + 6 t+ 9.

(a) Find the velocity at time t.

(b) Find the velocity at time t=3 seconds

Answer:

a) the velocity at time t is ( 15t² + 6 ) m/s

b) Velocity at time t=3 seconds is 141 m/s

Explanation:

Give the data in the question;

position of a particle is given by;

s = f(t) = 5t³ + 6t + 9

Velocity at t;

we differentiate with respect to t

so

V(t) = f'(t) = d/dt ( 5t³ + 6t + 9 )

V(t) = f(t) = 5(3t²) + 6(1) + 0 )

V(t) = f(t) = ( 15t²+6 ) m/s

Therefore, the velocity at time t is 15t²+6 m/s

b) Velocity at t = 3 seconds

V(t) = f(t) = ( 15t²+6 ) m/s

we substitute

V(3) = ( 15(3)² + 6 ) m/s

V(3) = ( (15 × 9) + 6 ) m/s

V(3) = ( 135 + 6 ) m/s

V(3) = 141 m/s

Therefore, Velocity at time t=3 is 141 m/s

3 0

2 years ago

A motorboat traveling 4 m/s, East encounters a current traveling 3.0 m/s, North. a) What is the resultant velocity of the motorb

Nostrana [21]

Explanation:

It is given that,

Velocity in East, $v_1=4\ m/s$

Velocity in North, $v_2=3\ m/s$

(a) The resultant velocity is given by :

$v=\sqrt{4^2+3^2}=5\ m/s$

(b) The width of the river is, d = 80 m

Let t is the time taken by the boat to travel shore to shore. So,

$t=\dfrac{d}{v}$

$t=\dfrac{80\ m}{5\ m/s}$

t = 16 seconds

(c) Let x is the distance covered by the boat to reach the opposite shore. So,

$x=v_2\times t$

$x=3\ m/s\times 16\ s$

x = 48 meters

Hence, this is the required solution.

4 0

3 years ago

Two equal positive charges q1 = q2 = 2μC are located at x = 0, y = 0.30 m and x = 0, y = -0.30 m, respectively. What are the mag

creativ13 [48]

Answer:

$Fx=486.58(-i)*10^{-2} N$ N: Resulting force on Q in the direction of the x axis

$Fy=162.49j(j)*10^{-2}$ N: Resulting force on Q in the direction of the y axis

Explanation:

To solve this problem we need 3 basic concepts:

1.Coulomb's law:

Two point charges (q1, q2) separated by a distance (r) exert a mutual force (F) whose magnitude is determined by the following formula:

$F=k*(\frac{q1*q2}{r^{2} } )$ Formula (1)

$K=8.99*10^{9} \frac{N*m^{2} }{C^{2} }$ : Coulomb constant

q 1,q2=charge in Coulombs (C)

The direction of F is determined by the unit vector ( $r_{u}$ ) in the direction of F

$r=r_{x} +r_{y}$ : position vector in the direction of F

r= (x2-x1)i+( y2-y1)j

The points (x1, y1) and (x2, y2) represent the position coordinates of the charges .

The sense of force depends on the sign of q1 and q2:

When the charges have opposite signs the force is attractive.

When the charges have an equal sign, the force is repulsion

2. Newton's third law: law of action and reaction

The magnitude of the force exerted by q1 on q2 is equal to the force exerted by q2 on q1, and these forces have the opposite direction.

<h3>F1-2=-F2-1 </h3>

Problem development

We have the forces exerted by the charges q1 + and q2 + on the load Q-.

The forces F1 and F2 on the load Q are attractive forces, Because of this the force comes out of Q and is directed towards q1 and q2

Calculation of vector r1 : points Q(0.4,0.15) , q1(0,0.3)

r1= (0- 0.4)i+(0.3-1.5)j

r1=-0.4i+0.15j

magnitude of r1

$r1=\sqrt{-0.4^{2}+0.15^{2} } =\sqrt{0.16+0.0225} =0.1825m$

Calculation of vector r2 : points(0.4,0.15) (0,-0.3)

r2= (0- 0.4)i+(-0.3-1.5)j

r2=(-0.4i-0.45j) m

magnitude of r2

$r2=\sqrt{-0.4^{2}+(-0.45)^{2} } =\sqrt{0.16+0.2025} =0.6 m$

We apply formula 1 to calculate the magnitudes of F1 and F2 with the known data :

q1 = q2 = 2μC , r1= 0.1825m ,r2=0.6m,

$K=8.99*10^{9} \frac{N*m^{2} }{C^{2} }$

1 μC= $10^{-6} C$

$F1=k*(2*10^{-6} *4*10^{-6} )/(0.1825^{2} )$

$F1=k*240.195*10^{-12}$

$F1=8.99*10^{9} *240.195*10^{-12}=2159.352*10^{-3} =215.93*10^{-2} N$

$F2=k*(2*10^{-6}*4*10^{-6}) /(0.6)^{2}$

$F2=8.99*10^{9} *22.2*10^{-12} =199.77*10^{-3} =19,97*10^{-2} N$

Calculation of the total electric force (F) exerted by the charges q1 and q2 on the charge Q

$r_{u}=(r_{x} +r_{y} )/r$ : unit vector in the direction of r

$F=F1(r_{u1})+F2(r_{u2} )$

ru=(rx+ry)/r

$F=\frac{215.93*(-0.4i+0.15j)}{0.1825} +\frac{19.93*(-0.4i-0.45j}{0.6} *10^{-2} N$

F=(-473.27i+177.47j-13.31i-14.98j) ))*10e-2 N

$F=(-486.58i+162.49j)*10^{-2} N$

Answer:

$Fx=486.58(-i)*10^{-2} N$ N: Resulting force on Q in the direction of the x axis

$Fy=162.49j(j)*10^{-2}$ N: Resulting force on Q in the direction of the y axis

8 0

3 years ago

Which piece of lab equipment would be least useful in finding the precise density of an unknown liquid?

miss Akunina [59]

D. Triple beam balance

3 0

3 years ago

Read 2 more answers