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3 years ago

13

A trip is taken that passes through the following points in order

1 answer:

MaRussiya [10]3 years ago

3 0

Answer:

40.0 m

...............

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Determining the pH of substances such as purple grape juice and catsup using test strips can be difficult. Why?

Charra [1.4K]

Answer:

Because they would naturally dye the test strips in the colors violet and red, regardless of their pH values

(would really appreciate the brainliest)

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3 years ago

What is the final step in the fourth stage of technological design

Ierofanga [76]

Answer:

after a product has been improved and approved? reporting the results finding ways to lower costs selling a prototype determining criteria.

Explanation:

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3 years ago

A ski jumper travels down a slope and

AleksandrR [38]

Answer:

304.86 metres

Explanation:

The x and y cordinates are $dcos\theta$ and $dsin\theta$ respectively

The horizontal distance travelled, $x=v_{ox}t=dcos\theta$

Making t the subject, $t=\frac{dcos\theta}{v_{ox}}$

Since $y=0.5gt^2=dsin\theta$ , we substitute t with the above and obtain

$0.5g(\frac{dcos\theta}{v_{ox}})^2=dsin\theta$

Making d the subject we obtain

$d=\frac{2v_{ox}^2sin\theta}{gcos^2\theta}$

$d=\frac{2*30^2sin48}{9.8cos^248}$

d=304.8584

d=304.86m

5 0

3 years ago

A hammer slides down a roof that makes a 30.0°angle with the horizontal. What are the magnitudes of the components of the hammer

dlinn [17]

<h2>Answer:7.14 $ms^{-1}$ ,4.125 $ms^{-1}$ </h2>

Explanation:

Whenever an object is moving in a 2D frame,its motion can be analysed as if it is travelling in two independent 1D frames.

One of such independent 1D frames are along horizontal and another along vertical.

Let $v$ be the total velocity.

Given that, $v=8.25ms^{-1}$

We call the horizontal velocity as $v_{h}$ and the vertical velocity as $v_{v}$ .

$v_{h}$ = $vCos\alpha$

$v_{v}=vSin\alpha$

where $\alpha$ is the angle between the object and horizontal.

It is given that $\alpha =30^{0}$

$v_{h}=8.25\times Cos(30^{0})=7.14ms^{-1}$

$v_{v}=8.25\times Sin(30^{0})=4.125ms^{-1}$

7 0

3 years ago

Consider a situation where a constant force of 25 N acts on an object having a mass of 2 kg for 3 seconds. What is the work done

blagie [28]

Answer:

Work done W =1406.25 J

Explanation:

Work done on a body can be calculated using newton's 2nd laws:

F=ma

$\Rightarrow a=\frac{F}{m}$

Hence acceleration of the block is given by:

$\Rightarrow a=\frac{25}{2}=12.5m/s^2$

Displacement of the object is given by:

$\Rightarrow S=ut+\frac{1}{2}at^2$

Substitute the values

$\Rightarrow S=0*3+\frac{1}{2}(12.5)3^2$

$\Rightarrow S=56.25 m$

Now work done is given by:

W=F.S

W = 25×56.25

W =1406.25 J

3 0

3 years ago