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3 years ago

from an origin you walk 3 meters east 7 meters west and then 6meters east . where are you now relative to the origin?

1 answer:

german3 years ago

3 0

You take 3 steps east, turn around & take another 7 steps west (4 steps always from origin) you turn around again and take 6 more steps east - 2 steps always from origin

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A force of 70 N is applied to a 28 kg rock causing it to slow down from 25 m/s to 15 m/s, a change in velocity of 10 m/s. How lo

maksim [4K]

Answer: 4 s

Explanation:

Given

The applied force is 70 N

mass of the rock is 28 kg

initial velocity $u=25\ m/s$

final velocity $v=15\ m/s$

Deceleration provided by force is

$a=-\dfrac{70}{28}=-2.5\ m/s^2$

using the equation of motion

$v=u+at\\\Rightarrow 15=25-2.5t\\\Rightarrow 2.5t=10\\\Rightarrow t=4\ s$

7 0

3 years ago

A car travels 20 km South, then turns and travels 30 km East? What is the total displacement?

Ne4ueva [31]

Answer:

Explanation:36.05 km

Given

First car travels $r_1=20\ km$ South

then turns and travels $r_2=30\ km$ east

Suppose south as negative y axis and east as positive x axis

So, $r_1=-20\hat{j}$

$r_2=30\hat{i}$

Displacement is the shortest between initial and final point

Dispalcement $=r=r_1+r_2$

Displacement $=-20\hat{j}+30\hat{i}$

Displacement $=30\hat{i}-20\hat{j}$

Magnitude $=\sqrt{30^2+(-20)^2}$

Magnitude $=36.05\ km$

4 0

3 years ago

What are the types of energies released when a candle burns?

Phantasy [73]

When a candle is burning the candle is releasing thermal and radiant energy

4 0

3 years ago

Read 2 more answers

Use your understanding of heat loss to ESTIMATE the cost of the lost energy through one standard window during an average summer

miskamm [114]

Answer:

The cost of energy is $ 0.34.

Explanation:

The energy is the capacity to do work.

The energy is a scalar quantity and its SI unit is Joule.

The commercial unit of energy is kWh.

Cost of 1 kWh energy = $ 0.17

energy loss by standard window is 2 kWh .

So, the cost of lost of energy is

Cost = $ 0.17 x 2 = $ 0.34

3 0

3 years ago

An object with mass 3.5 kg is attached to a spring with spring stiffness constant k = 270 N/m and is executing simple harmonic m

Elanso [62]

Answer:

Part a)

A = 0.066 m

Part b)

maximum speed = 0.58 m/s

Explanation:

As we know that angular frequency of spring block system is given as

$\omega = \sqrt{\frac{k}{m}}$

here we know

m = 3.5 kg

k = 270 N/m

now we have

$\omega = \sqrt{\frac{270}{3.5}}$

$\omega = 8.78 rad/s$

Part a)

Speed of SHM at distance x = 0.020 m from its equilibrium position is given as

$v = \omega \sqrt{A^2 - x^2}$

$0.55 = 8.78 \sqrt{A^2 - 0.020^2}$

$A = 0.066 m$

Part b)

Maximum speed of SHM at its mean position is given as

$v_{max} = A\omega$

$v_{max} = 0.066(8.78) = 0.58 m/s$

4 0

3 years ago