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anyanavicka [17]

2 years ago

7

The releasing of an object from certain height to ground due to gravity is called____ A, Free falling B, Projectile C, curved li

near D,para pola

1 answer:

MArishka [77]2 years ago

7 0

Answer:

projectile

Explanation:

when an object is released from certain height to ground due to gravity is called projectile

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Space scientists have a large test chamber from which all the air can be evacuated and in which they can create a horizontal uni

nexus9112 [7]

Answer:

the magnitude of the electric force on the projectile is 0.0335N

Explanation:

time of flight t = 2·V·sinθ/g

= (2 * 6.0m/s * sin35º) / 9.8m/s²

= 0.702 s

The body travels for this much time and cover horizontal displacement x from the point of lunch

So, use kinematic equation for horizontal motion

horizontal displacement

x = Vcosθ*t + ½at²

2.9 m = 6.0m/s * cos35º * 0.702s + ½a * (0.702s)²

a = -2.23 m/s²

This is the horizontal acceleration of the object.

Since the object is subject to only electric force in horizontal direction, this acceleration is due to electric force only

Therefore,the magnitude of the electric force on the projectile will be

F = m*|a|

= 0.015kg * 2.23m/s²

= 0.0335 N

Thus, the magnitude of the electric force on the projectile is 0.0335N

3 0

2 years ago

Read 2 more answers

An arch carries the thrust of weight to its _____(1)______. With a _____(2)______, the horizontal part of the structure supports

finlep [7]

Option D is correct. An arch carries the thrust of weight to its <u>sides </u>with a <u>post-and-lintel.</u>

<u></u>

<h3>What is an arch?</h3>

An arch is indeed a vertical curving construction that covers an elevated space that may or may not sustain the load above it or the pressure gradient against it

In the case of a horizontally arched, such as an embankment dam. While arches and vaults are often confused, A vault is defined as an ongoing arch forming a roof.

Option D satisfies the fill-in blanks option.

Hence option D is correct. An arch carries the thrust of weight to its <u>sides </u>with a <u>post-and-lintel.</u>

<u></u>

To learn more about the arch refer to the link;

brainly.com/question/18162421

6 0

2 years ago

Given the vector A with components Ax = 2.00, Ay = 6.00, the vector B with components Bx = 2.00, By = 22.00, and the vector D =

nekit [7.7K]

Answer:

<em>The magnitude of vector d is 16 and the angle with the x-axis is 270°</em>

Explanation:

<u>Operations With Vectors</u>

Given two vectors in rectangular components:

$\vec a=(ax,ay)\ ,\ \vec b=(bx,by)$

The sum of the vectors is:

$\vec a+\vec b=(ax+bx,ay+by)$

The difference between the vectors is:

$\vec a-\vec b=(ax-bx,ay-by)$

The magnitude of $\vec a$ is:

$|\vec a|=\sqrt{ax^2+ay^2}$

The angle $\vec a$ makes with the horizontal positive direction is:

$\displaystyle \tan\theta=\frac{ay}{ax}\\$

The question provides the vectors:

$\vec a=(2,6)$

$\vec b=(2,22)$

$\vec d=\vec a-\vec b$

Calculate:

$\vec d=(2,6)-(2,22)=(0,-16)$

The magnitude of $\vec d$ is:

$|\vec d|=\sqrt{0^2+(-16)^2}=\sqrt{0+256}=16$

The angle is calculated by:

$\displaystyle \tan\theta=\frac{-16}{0}$

The division cannot be calculated because the denominator is zero. We need to estimate the correct angle by looking at the components of the vector. Since the x-coordinate is zero and the y-coordinate is negative, the vector points downwards (south), thus the angle must be -90° or 270° if the range goes from 0° to 360°.

The magnitude of vector d is 16 and the angle with the x-axis is 270°

4 0

2 years ago

What is it that if you have, you want to share me, and if you share, you do not have?

Colt1911 [192]

Answer:

Secrets? Cuz if I share you the secrets then the secrets will no longer will be secrets.

4 0

2 years ago

Read 2 more answers

if pete (mass =90.00kg) weighs himself and finds that he weighs 30.0 pounds, how far away from the earth is he?

Svet_ta [14]

Here we know that mass of the person is 90 kg

His weight is given as 30 lbf

so here we can convert it into Newton as we know that

1 lbf = 4.45 N

Now from above conversion

$30 lbf = 30 \times 4.45 = 133.45 N$

now we can use this to find the gravity at this height

$mg = 133.45$

$90\times g = 133.45$

$g = 1.49 m/s^2$

now we know that with height gravity varies as

$g' = g(\frac{R}{R+H})^2$

$1.49 = 9.8(\frac{6.37\times 10^6}{6.37\times 10^6 + h})^2$

$h = 1.0 \times 10^7 m$

so above is the height from the surface of earth

4 0

2 years ago