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2 years ago

Michael is 4 feet tall. Explain how Michael could find his height in inches. Then explain how he could find his height in yards.

1 answer:

6 0

Well he could take his height 4 feet and multiply it by 12 to get his height in inches which is 48 inches then to find his height in yards he would take his height 4 feet and divide it by 3 to get 1.333 repeating. hope this helps

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On a snowy day, max (mass = 15 kg) pulls his little sister maya in a sled (combined mass = 20 kg) through the slippery snow. max

sesenic [268]

Work done by a given force is given by

$W = F.d$

here on sled two forces will do work

1. Applied force by Max

2. Frictional force due to ground

Now by force diagram of sled we can see the angle of force and displacement

work done by Max = $W_1 = Fdcos\theta$

$W_1 = 12*5cos15$

$W_1 = 57.96 J$

Now similarly work done by frictional force

$W_2 = Fdcos\theta$

$W_2 = 4*5cos180$

$W_2= -20 J$

Now total work done on sled

$W_{net}= W_1 + W2$

$W_{net} = 57.96 - 20 = 37.96 J$

7 0

2 years ago

Please help! will give brainlest!!!!!!!!!!!!

eimsori [14]

The force of friction is 19.1 N

Explanation:

According to Newton's second law, the net force acting on the bag is equal to the product between its mass and its acceleration:

$\sum F = ma$

where

$\sum F$ is the net force

m is the mass

a is the acceleration

The bag is moving at constant speed, so its acceleration is zero:

$a=0$

Therefore the net force is zero as well:

$\sum F = 0$

Here we are interested only in the forces acting along the horizontal direction, therefore the net force is given by:

$\sum F = F cos \theta - F_f = 0$

where

$F cos \theta$ is the horizontal component of the applied force, with

F = 22.5 N

$\theta=32.0^{\circ}$

$F_f$ is the force of friction

And solving for $F_f$ , we find

$F_f =Fcos \theta=(22.5)(cos 32.0^{\circ})=19.1 N$

Learn more about friction:

brainly.com/question/6217246

brainly.com/question/5884009

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#LearnwithBrainly

7 0

2 years ago

The magnitude J(r) of the current density in a certain cylindrical wire is given as a function of radial distance from the cente

kipiarov [429]

Answer:

$I=68.31\times 10^{-6}\ A$

Explanation:

Given that

J(r) = Br

We know that area of small element

dA = 2 π dr

I = J A

dI = J dA

Now by putting the values

dI = B r . 2 π dr

dI= 2π Br² dr

Now by integrating above equation

$\int_{0}^{I}dI= \int_{r_1}^{r_2}2\pi Br^2 dr$

$I={2\pi B}\times \dfrac{r_2^3-r_1^3}{3}$

Given that

B= 2.35 x 10⁵ A/m³

r₁ = 2 mm

r₂ = 2+ 0.0115 mm

r₂ = 2.0115 mm

$I={2\pi B}\times \dfrac{r_2^3-r_1^3}{3}$

By putting the values

$I={2\pi \times 2.35 \times 10^5 }\times \dfrac{(2.0115\times 10^{-3})^3-(2\times 10^{-3})^3}{3}\ A$

$I=68.31\times 10^{-6}\ A$

7 0

2 years ago

Read 2 more answers

A fireboat is to fight fires at coastal areas by drawing seawater with a density of 1030 kg/m3 through a 10-cm-diameter pipe at

GaryK [48]

Answer:

50.93 m/s

199.5 kW

Explanation:

From the question, the nozzle exit diameter = 5 cm, Radius= diameter/2= 5cm/2= 2.5cm. we can convert it to metre for unit consistency= (2.5×0.01)=

0.025m

We can calculate the The cross sectional area of the nozzle as

A= πr^2

A= π ×0.025^2

= 1.9635 ×10^- ³ m²

From the question, the water is moving through the pipe at a rate of 0.1 m /s , then for the water to move through it at a seconds, it must move at

(0.1 / 1.9635 ×10^- ³ m²)

= 50.93 m/s

During the Operation of the pump, the Dynamic energy of the water= potential energy provided there is no loss during the Operation

mgh = 1/2mv²

We can make "h" subject of the formula, which is the height of required head of water

h = (1/2mv²)/mg

h= v² / 2g

h = 50.93² / (2 ×9.81)

h = 132.21m

From the question;

The total irreversible head loss of the system = 3 m,

the given position of nozzle = 3 m

the total head the pump needed=(The total irreversible head loss of the system + the position of the nozzle + required head of water )

=(3 + 3 + 132.21m)

=138.21m

mass of water pumped in a seconds can be calculated since we know that mass is a product of volume and density

Volume= 0.1m³

Density of sea water=1030 kg/m

(0.1 m^3× 1030)

= 103kg

We can calculate the Potential enegry, which is = mgh

= (103 ×9.81 × 138.21)

= 139651.5 Watts

= 139.65kW

To determine required shaft power input to the pump and the water discharge velocity

Energy= efficiency × power

But we are given efficiency of 70 percent, then

139651.5 Watts = 0.7P

=199502.18 Watts

P=199.5 kW

Therefore, the required shaft power input to the pump and the water discharge velocity is 199.5 kW

5 0

1 year ago

Students use a stretched elastic band to launch carts of known mass horizontally on a track. The elastic bands exert a force F,

rodikova [14]

Answer:

the correct answer is E

A graph of the cart's maximum speed squared as a function of x^3

Explanation:

For this exercise let's use Newton's second law

F = m a

force has the form

F = k x²

and acceleration is related to velocity

a = dv / dt

Let's use the chain rule or L'Hospital

a = dv /dx dx/dt

a = dv /dx v

let's substitute

k x² = m v dv / dx

k /m x² dx = v dv

we integrate

k /m x³ /3 = v² / 2

v² = (2k /3m) x³

This is the expression for the variation of the speed as a function of the position, to make a linear graph realism the changes of variable

y = v²

x´ = x³

y = (2k/3m) x´

if we graph y vs x 'we have a linear graph whose slope is

m = 2k / 3m

By reviewing the different answers, the correct answer is E

4 0

2 years ago