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Naily [24]
3 years ago
13

A runner wants to run 12.0 km . Her running pace is 8.2 mi/hr . How many minutes must she run? Express your answer using two sig

nificant figures.
Physics
1 answer:
Hunter-Best [27]3 years ago
5 0

Answer:

53 minutes

Explanation:

To solve this problem, we need to use common proportions to convert.

= 11.7km * 1mi/1.61km * 1hr/8.2mi * 60min/hr

≈ 53 minutes

Best of LUck!

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If the angle of reflection is 25 degrees, the angle of incidence is
natali 33 [55]

Answer:

25 degrees

Explanation:

The angle of incidence equals the angle of reflection

7 0
3 years ago
I NEED HELP PLEASE, THANKS! :)
mrs_skeptik [129]

Answer:

1. Largest force: C;  smallest force: B; 2. ratio = 9:1

Explanation:

The formula for the force exerted between two charges is

F=K\dfrac{ q_{1}q_{2}}{r^{2}}

where K is the Coulomb constant.

q₁ and q₂ are also identical and constant, so Kq₁q₂ is also constant.

For simplicity, let's combine Kq₁q₂ into a single constant, k.

Then, we can write  

F=\dfrac{k}{r^{2}}

1. Net force on each particle

Let's

  • Call the distance between adjacent charges d.
  • Remember that like charges repel and unlike charges attract.

Define forces exerted to the right as positive and those to the left as negative.

(a) Force on A

\begin{array}{rcl}F_{A} & = & F_{B} + F_{C} + F_{D}\\& = & -\dfrac{k}{d^{2}}  - \dfrac{k}{(2d)^{2}}  +\dfrac{k}{(3d)^{2}}\\& = & \dfrac{k}{d^{2}}\left(-1 - \dfrac{1}{4} + \dfrac{1}{9} \right)\\\\& = & \dfrac{k}{d^{2}}\left(\dfrac{-36 - 9 + 4}{36} \right)\\\\& = & \mathbf{-\dfrac{41}{36} \dfrac{k}{d^{2}}}\\\\\end{array}

(b) Force on B

\begin{array}{rcl}F_{B} & = & F_{A} + F_{C} + F_{D}\\& = & \dfrac{k}{d^{2}}  - \dfrac{k}{d^{2}}  + \dfrac{k}{(2d)^{2}}\\& = & \dfrac{k}{d^{2}}\left(\dfrac{1}{4} \right)\\\\& = &\mathbf{\dfrac{1}{4} \dfrac{k}{d^{2}}}\\\\\end{array}

(C) Force on C

\begin{array}{rcl}F_{C} & = & F_{A} + F_{B} + F_{D}\\& = & \dfrac{k}{(2d)^{2}} + \dfrac{k}{d^{2}}  + \dfrac{k}{d^{2}}\\& = & \dfrac{k}{d^{2}}\left( \dfrac{1}{4} +1 + 1 \right)\\\\& = & \dfrac{k}{d^{2}}\left(\dfrac{1 + 4 + 4}{4} \right)\\\\& = & \mathbf{\dfrac{9}{4} \dfrac{k}{d^{2}}}\\\\\end{array}

(d) Force on D

\begin{array}{rcl}F_{D} & = & F_{A} + F_{B} + F_{C}\\& = & -\dfrac{k}{(3d)^{2}}  - \dfrac{k}{(2d)^{2}}  - \dfrac{k}{d^{2}}\\& = & \dfrac{k}{d^{2}}\left( -\dfrac{1}{9} - \dfrac{1}{4} -1 \right)\\\\& = & \dfrac{k}{d^{2}}\left(\dfrac{-4 - 9 -36}{36} \right)\\\\& = & \mathbf{-\dfrac{49}{36} \dfrac{k}{d^{2}}}\\\\\end{array}

(e) Relative net forces

In comparing net forces, we are interested in their magnitude, not their direction (sign), so we use their absolute values.

F_{A} : F_{B} : F_{C} : F_{D}  =  \dfrac{41}{36} : \dfrac{1}{4} : \dfrac{9}{4} : \dfrac{49}{36}\ = 41 : 9 : 81 : 49\\\\\text{C experiences the largest net force.}\\\text{B experiences the smallest net force.}\\

2. Ratio of largest force to smallest

\dfrac{ F_{C}}{ F_{B}} = \dfrac{81}{9} = \mathbf{9:1}\\\\\text{The ratio of the largest force to the smallest is $\large \boxed{\mathbf{9:1}}$}

7 0
3 years ago
If this decay has a half-life of 10.2 years, what mass of 60.8g carbon-14 will remain after 20.4 years
OLEGan [10]
20.4 years is 20.4/10.2 = 2 half-life cycles, which means a quarter of the starting mass or 15.2 g will remain after this time.
5 0
3 years ago
Explain why, in terms of forces, there is a risk of head injury when diving from this height. Suggest why the high divers would
podryga [215]

Answer:

Due to lower risk of injury or damage.

Explanation:

The high divers would choose to enter the water from the feet first because there is low risk of injury. The brain is the most important part of the body which very sensitive to any small injury. Small injury to brain leads to big problems in life. High divers can reach speeds of nearly 60 mph and enters about 28m into the water in about three seconds which can damage the head region if comes in contact with the ground so this is the reason the high divers avoid of entering in the water through their heads and choose entering through their feet.

4 0
3 years ago
How many electrons flow through a point in a wire in 7.00 s if there is a constant current of I = 4.35 A?
OLEGan [10]

Answer:

1.90×10²⁰ Electrons

Explanation:

From the question,

Q = It.................... Equation 1

Where Q = charge flowing through the wire, I = current, t = time

Given: I = 4.35 A, t = 7.00 s

Substitute these values into equation 1

Q = 4.35(7.00)

Q = 30.45 C.

But,

1 electron contains 1.6×10⁻¹⁹ C

therefore,

30.45 C = 30.45/1.6×10⁻¹⁹  electrons

= 1.90×10²⁰ Electrons

8 0
3 years ago
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