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3 years ago

Kenna has a mass of 30 kilograms. What's her weight in newtons? Assume that acceleration due to gravity is 9.8 N/kg.

Physics

1 answer:

astra-53 [7]3 years ago

5 0

Answer:

294 N

Explanation:

The weight of an object is the gravitational force exerted by the Earth on the object.

The direction of this force is always towards the Earth's centre, while the magnitude is given by:

$W=mg$

where

W is the weight

m is the mass of the object

g is the acceleration due to gravity

Here we have:

m = 30 kg is the mass of the object

g = 9.8 N/kg is the acceleration due to gravity

So, the weight of the object is

$W=(30)(9.8)=294 N$

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erastovalidia [21]

Answer:

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2 years ago

A constant net force of 20 N is applied to a 32kg car, causing it to speed up from 4.0 to 9.0 m/s. How long is the force applied

pishuonlain [190]

Answer:

8 seconds

Explanation:

From Newton's second law;

Ft = m(v-u)

F = Force applied

t = time taken

v = final velocity

u = initial velocity

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20t = 32 * 5

t = 32 * 5/ 20

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8 0

2 years ago

Using Excel, or some other graphing software, plot the values of y as a function of x. (You will not submit this spreadsheet. Ho

Evgesh-ka [11]

Answer:

a) > x<-c(1,2,3,4,5)

> y<-c(1.9,3.5,3.7,5.1,6)

> linearmodel<-lm(y~x)

And the output is given by:

> linearmodel

Call:

lm(formula = y ~ x)

Coefficients:

(Intercept) x

1.10 0.98

b) $y = 0.98 x +1.10$

And if we compare this with the general model $y = mx +b$

We see that the slope is m= 0.98 and the intercept b = 1.10

Explanation:

Part a

For this case we have the following data:

x: 1,2,3,4,5

y: 1.9,3.5,3.7,5.1, 6

For this case we can use the following R code:

> x<-c(1,2,3,4,5)

> y<-c(1.9,3.5,3.7,5.1,6)

> linearmodel<-lm(y~x)

And the output is given by:

> linearmodel

Call:

lm(formula = y ~ x)

Coefficients:

(Intercept) x

1.10 0.98

Part b

For this case we have the following trend equation given:

$y = 0.98 x +1.10$

And if we compare this with the general model $y = mx +b$

We see that the slope is m= 0.98 and the intercept b = 1.10

7 0

3 years ago

Ignoring air resistance and the little friction from the plastic tube, the magnet was a freely-falling object in each trial. If

horsena [70]

Answer:

emf will also be 10 times less as compared to when it has fallen $40 \mathrm{m}$

Explanation:

We know, from faraday's law-

$e m f=-N \frac{\Delta \Phi}{\Delta T}$

and $\Phi=B . A$

So, as the height increases the velocity with which it will cross the ring will also increase. $(v=\sqrt{2 g h})$

Given

$\mathrm{V} 1(\text { Speed at } 40 \mathrm{m})=2 \mathrm{x} \mathrm{V} 2(\text { speed at } 10 \mathrm{m})$

$\sqrt{2 g h_{2}}=2 \times \sqrt{2 g h_{1}}=28.28 \mathrm{m} / \mathrm{s}$

Now, from $40 \mathrm{cm}$

$V_{3}=\sqrt{2 g h_{3}}=\sqrt{2 \times 10 \times 0.4}=2.82 \mathrm{m} / \mathrm{s}$

From equation a and b we see that velocity when dropped from $40 \mathrm{m}$ is 10 times greater when height is 40 $\mathrm{cm}$ so, emf will also be 10 times less as compared to when it has fallen $40 \mathrm{m}$

4 0

2 years ago

15 points for 2 questions

Elden [556K]

I believe the answer would be c because i think that you multiply the 2

8 0

2 years ago

Read 2 more answers