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MariettaO [177]

3 years ago

14

Monique walks 360 meters to get to lanier on days when she is early and doesn't get caught by traffic it takes her 60 seconds to

get to school how fast was she running

1 answer:

Mars2501 [29]3 years ago

4 0

Answer:

6m/s

Explanation:

We are to calculate the speed of Monique

Speed = Distance/Time

Given

Distance = 360m

Time = 60secs

Substitute

Speed = 360m/60s

Soeed = 6m/s

Hence she was running at 6m/s

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Two blocks are connected by a massless rope that passes over a 1 kg pulley with a radius of 12 cm. The rope moves over the pulle

tangare [24]

Answer:

$F=1.159$

Explanation:

From the question we are told that:

Mass of pulley $M=1kg$

Radius $r=12cm$

Mass of block A $M_a=2.1kg$

Mass of block B $m_b=4.1kg$

Spring constant $\mu= 358 J/m2$

Generally the equation for Torque is mathematically given by

Since $\sumF=ma$

At mass A

$T_2-f_3=2.1a$

At mass B

$4.8-T_1=4.1a$

At Pulley

$R(T_1-T_2)=\frac{1*1*R^2}{2}\frac{a}{R}$

$R(T_1-T_2)=0.55a$

Therefore the equation for total force F

At mass A+At mass B+At Pulley

$(T_2-f_3+4.8-T_1+R(T_1-T_2)=2.1a+4.1a+0.55a$

$(T_2-f_3+4.8-T_1+R(T_1-T_2)=2.7a+4.8a+0.55a$

$-f_3+4.1=6.75a$

$-f_3=6.75a+4.8$

Since From above equation

$M_{eff}=6.7kg$

Therefore

$T=2\pi \sqrt{{\frac{M_{eff}}{k}}$

$T=2\pi \sqrt{{\frac{6.75}{\mu}}$

$T=0.862s$

Generally the equation for frequency is mathematically given by

$F=\frac{1}{T} \\F=\frac{1}{0.862}$

$F=1.159$

3 0

3 years ago

What SI unit is used to measure mechanical energy

VashaNatasha [74]

The SI unit of energy is the Joule .

Any kind of energy ... electrical, mechanical, nuclear, solar,
wind etc. It's easy to change from one form to another ... we
do it every day ... and they all have the same unit.

7 0

4 years ago

Read 2 more answers

A string is attached to a ball that has a mass of 0.11 kg. A student pulls up on the string so that the ball accelerates upward

enot [183]

Answer:

T=+1.133N

Explanation:

Tension and weight are forces that have opposite directions

Weight is negative (downward)

W=m*g= 0.11kg*(-9.8m/s^2)

W= -1.078N

Tension is possitive (upward)

The total force will be the sum of both (the difference taking in consideration the direction)

Ft= T+W

Also the total force is the product of the mass due to acceleration:

Ft=m*a

Ft= +0.11kg*0.5m/s^2

Ft=+0.055N (upward)

Tension will be the difference between Ft and W:

T= Ft-W

T=+0.055N-(-1.078N)

T=+1.133N

7 0

3 years ago

Use the SI prefixes in Table 3 of this chapter to convert these hypothetical units of measure into appropriate quantities: a. 10

Amanda [17]

<span>The first response would be "deca," since this is a multiple of 10^1. B, since it's working off the thousands prefix (10^3), would be "kilo." The third, at 10^-6, would be "micro." Next, at 10^-9, would be "nano," and the final, 10^18, would have a prefix of "exa."</span>

6 0

3 years ago

Four forces act on bolt A as shown; F1 150N, F2 80N, F3 110N and F4 100N. Determine the magnitude and direction of the resultant

netineya [11]

Complete Question

The complete question(reference (chegg)) is shown on the first uploaded image

Answer:

The magnitude of the resultant force is $F = 199.64 \ N$

The direction of the resultant force is $\theta = 4.1075^o$ from the horizontal plane

Explanation:

Generally when resolving force, if the force (F )is moving toward the angle then the resolve force will be $Fcos(\theta )$ while if the force is moving away from the angle then the resolved force is $Fsin (\theta )$

Now from the diagram let resolve the forces to their horizontal component

So

$\sum F_x = 150 cos(30) + 100cos(15) -80sin (20)$

$\sum F_x = 199.128 \ N$

Now resolving these force into their vertical component can be mathematically evaluated as

$\sum F_{y} = 150 sin(30) - 100sin(15) -110 +80 cos(20)$

$\sum F_{y} = 14.30$

Now the resultant force is mathematically evaluated as

$F = \sqrt{F_x^2 + F_y^2}$

substituting values

$F = \sqrt{199.128^2 + 14.3^2}$

$F = 199.64 \ N$

The direction of the resultant force is evaluated as

$\theta = tan^{-1}[\frac{F_y}{F_x} ]$

substituting values

$\theta = tan^{-1}[\frac{ 14.3}{199.128} ]$

$\theta = 4.1075^o$ from the horizontal plane

5 0

4 years ago