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zvonat [6]
3 years ago
15

If you run at 8m/s for 15 minutes how far will you go

Physics
2 answers:
enyata [817]3 years ago
8 0

Answer:

<h2>hope it helps you see the attachment for further information ......✌✌✌✌✌</h2>

sashaice [31]3 years ago
6 0

Answer: 7200 m

Explanation: The solution is, first convert 15 minutes to seconds.

15 mins x 60 s / 1 min = 900 s

Use the formula for speed which is v= d/t then derive for d.

d = vt

= 8 m/s ( 900s)

= 7200 m

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Tornado- Trees knocked down, debris everywhere, ground and dirt scattered.
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3 years ago
If a 1.00 kg body has an acceleration of 2.44 m/s2 at 53° to the positive direction of the x axis, then what are (a) the x comp
Ilia_Sergeevich [38]

(a) Fx = 1.464 N

(b) Fy = 1.952 N

(c) F(x, y) = 1.464 i + 1.952 j

Given

Mass = 1kg

Acceleration = 2.44 m/s2

Angle with positive X axis = 53°

As we know

F = ma

By substituting value

F= 1×2.44 N

F= 2.44 N

(a)   Component of force in X direction

Fx = F Cosθ

Fx = 2.44 Cos(53°)

Fx = 2.44 × 0.60 = 1.464 N

(b) Component of force in Y direction

Fy = F Sinθ

Fy = 2.44 Sin(53°) = 2.44 × 0.80 = 1.952 N

(c) Net force in vector notation

F(x, y) = 1.464 i + 1.952 j

Thus we got net force.

#SPJ4

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6 0
2 years ago
The scientist who suggested that energy can be created under certain conditions was
tester [92]
Einstein hope this helped
7 0
3 years ago
Read 2 more answers
A point charge with charge q1 = 3.40 μC is held stationary at the origin. A second point charge with charge q2 = -4.90 μC moves
expeople1 [14]

Answer:

-0.79 J

Explanation:

We are given that

q_1=3.4\mu C=3.4\times 10^{-6} C

1\mu C=10^{-6} C

q_2=-4.9\mu C=-4.9\times 10^{-6} C

x_1=0.125,y_1=0

x_2=0.280,y_2=0.235

We have to find the work done by the electric force on the moving point charge.

r_1=\sqrt{x^2_1+y^2_1}=\sqrt{(0.125)^2+0}=0.125

r_2=\sqrt{(0.280)^2+(0.235)^2}=0.366

Work done,W=kq_1q_2(\frac{1}{r_1}-\frac{1}{r_2})

Where k=9\times 10^9

Using the formula

W=9\times 10^9\times 3.4\times 10^{-6}\times(-4.9\times 10^{-6})(\frac{1}{0.125}-\frac{1}{0.366})

W=-0.79 J

5 0
4 years ago
Suppose a rectangular piece of aluminum has a length D, and its square cross section has the dimensions W XW, where D (W x W) to
Ludmilka [50]

Answer:

R₂ / R₁ = D / L

Explanation:

The resistance of a metal is

        R = ρ L / A

Where ρ is the resistivity of aluminum, L is the length of the resistance and A its cross section

We apply this formal to both configurations

Small face measurements (W W)

The length is

         L = W

Area  

         A = W W = W²

        R₁ = ρ W / W² = ρ / W

Large face measurements (D L)

       Length L = D= 2W

       Area     A = W L

     R₂ = ρ D / WL = ρ 2W / W L = 2 ρ/L

The relationship is

    R₂ / R₁ = 2W²/L

6 0
3 years ago
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