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Dovator [93]
3 years ago
15

You are asked to design an experiment which is similar to Match Graph activity. If you move 3m back from the motion sensor in 2s

. What is your speed?
Physics
1 answer:
Art [367]3 years ago
5 0

Answer:

The speed is 1.5 m/s.

Explanation:

Given that,

Distance = 3 m

Time = 2 sec

We know that,

The position and time graph shows the speed of motion.

We need to calculate the speed

Using formula of speed

v=\dfrac{d}{t}

Where, d = distance

t = time

Put the value into the formula

v=\dfrac{3}{2}

v=1.5\ m/s

Hence, The speed is 1.5 m/s.

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A cyclist is riding his bike up a mountain trail. When he starts up the trail, he is going 8 m/s. As the trail gets steeper, he
shusha [124]

Answer:

a)   a = - 0.0833 m / s²,  b)   t = 4.4 s

Explanation:

a) this is a kinematics exercise where the acceleration is along the inclined plane

         v = v₀ - a t

         a = v₀ - v / t

         a = 3 - 8/60

         a = - 0.0833 m / s²

b) in this case the final velocity is zero

         v = v₀ - a t

         0 = v₀ - at

         t = v₀ / a

         t = 28 / 6.4

         t = 4.375 s

         t = 4.4 s

3 0
3 years ago
________ is a force acting through distance.
Andreas93 [3]
High school???
No way
It's work.
8 0
3 years ago
An electron that has an instantaneous velocity of ???? = 2.0 × 106 m ???? ???? + 3.0 × 106 m ???? ???? is moving through the uni
Setler79 [48]

Explanation:

It is given that,

Velocity of the electron, v=(2\times 10^6i+3\times 10^6j)\ m/s

Magnetic field, B=(0.030i-0.15j)\ T

Charge of electron, q_e=-1.6\times 10^{-19}\ C

(a) Let F_e is the force on the electron due to the magnetic field. The magnetic force acting on it is given by :

F_e=q_e(v\times B)

F_e=1.6\times 10^{-19}\times [(2\times 10^6i+3\times 10^6j)\times (0.030i-0.15j)]

F_e=-1.6\times 10^{-19}\times (-390000)(k)

F_e=6.24\times 10^{-14}k\ N

(b) The charge of electron, q_p=1.6\times 10^{-19}\ C

The force acting on the proton is same as force on electron but in opposite direction i.e (-k). Hence, this is the required solution.

8 0
3 years ago
What is the magnitude of the force needed to keep a 60 newton rubber block moving across level,dry asphalt in a straight line at
pantera1 [17]

let Coefficients of Friction of Rubber on asphalt (dry) =0.7

F= Coefficients of Friction * normal force = 0.7 * 60 =42 N

so the net force of the rubber is zero, meaning it will travel at a constant speed.

When the rubber is travel at 2m/s, 42N is required to keep moving at constant speed

6 0
4 years ago
Read 2 more answers
You are planning to make an open rectangular box from an 8-inch by 15-inch piece of cardboard by cutting congruent squares from
Phoenix [80]

Let us say that x is the cut that we will make on the sides to make a box, therefore the new dimensions are:

l = 15 – 2x

w = 8 – 2x

It is 2x since we cut on two sides.

 

We know that volume is:

V = l w x

V = (15 – 2x) (8 – 2x) x

V = 120x – 30x^2 – 16x^2 + 4x^3

V = 120x – 46x^2 + 4x^3

 

Taking the 1st derivative:

dV/dx = 120 – 92x + 12x^2

 

Set dV/dx = 0 to get maxima:

120 – 92x + 12x^2 = 0

 

Divide by 12:

x^2 – (92/12)x + 10 = 0

(x – (92/24))^2 = -10 + (92/24)^2

x - 92/24 = ±2.17

x = 1.66, 6

We cannot have x = 6 because that will make our w negative, so:

x = 1.66 inches

 

So the largest volume is:

V = 120x – 46x^2 + 4x^3

V = 120(1.66) – 46(1.66)^2 + 4(1.66)^3

V = 90.74 cubic inches

4 0
4 years ago
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