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mestny [16]
2 years ago
8

PLEASE HELP ME WILL GIVE BRAINLIEST

Physics
2 answers:
weqwewe [10]2 years ago
7 0

Answer:

A.always changing

Explanation:

Mekhanik [1.2K]2 years ago
7 0

Answer:

always equal

The horizontal and vertical components of a projectile's velocity is always equal. The horizontal and vertical have different charges.

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If an airplane is flying at 300 km/h to the east and is facing a headwind of 18.0 km/h, what is the airplane's final velocity?
givi [52]
If an airplane is flying at 300 km/h to the east and is facing a headwind of 18.0 km/h, the final velocity can be calculated using simple vector addition. In this case, the planes velocity is positive (+330 km/h) and head wind has a negative component (-18.0 km/h). Vector addition yields +330 km / h + (-18.0 km /h) = 312 km / h. 
8 0
3 years ago
A charged particle having mass 6.64 x 10-27 kg (that of a helium atom) moving at 8.70 x 105 m/s perpendicular to a 1.30-T magnet
Fiesta28 [93]

Answer:

the charge of the particle is 2.47 x 10⁻¹⁹ C

Explanation:

Given;

mass of the particle, m = 6.64 x 10⁻²⁷ kg

velocity of the particle, v = 8.7 x 10⁵ m/s

strength of the magnetic field, B = 1.3 T

radius of the circle, r = 18 mm = 1.8 x 10⁻³ m

The magnetic force experienced by the charge is calculated as;

F = ma = qvB

where;

q is the charge of the particle

a is the acceleration of the charge in the circular path

a = \frac{v^2}{r} \\\\ma = qvB\\\\q = \frac{ma}{vB} \\\\q = \frac{mv^2}{rvB} = \frac{mv}{rB} \\\\q = \frac{(6.64\times 10^{-27} ) \times (8.7\times 10^5)}{(1.8\times 10^{-2}) \times (1.3)} \\\\q = 2.47 \ \times 10^{-19} \ C

Therefore, the charge of the particle is 2.47 x 10⁻¹⁹ C

6 0
3 years ago
A single mass (m1 = 3.5 kg) hangs from a spring in a motionless elevator. The spring constant is k = 278 N/m. 1)What is the dist
artcher [175]
<h2>Answer:</h2>

0.126m

<h2>Explanation:</h2>

According to Hooke's law, the force (F) acting on a spring to cause an extension or compression (e) is given by;

F = k x e            -------------------(i)

Where;

k = the spring's constant.

From the question, the force acting on the spring is the weight(W) of the mass. i.e

F = W               -----------------------(ii)

<em>But;</em>

W = m x g;

where;

m = mass of the object

g = acceleration due to gravity [usually taken as 10m/s²]

<em>From equation (ii), it implies that;</em>

F = W = m x g

<em>Now substitute F = m x g into equation(i) as follows;</em>

F = k x e

m x g = k x e      ------------------(iii)

<em>From the question;</em>

m = m1 = 3.5kg

k = 278N/m

<em>Substitute these values into equation (iii) as follows;</em>

3.5 x 10 = 278 x e

35 = 278e

<em>Now solve for e;</em>

e = 35/278

e = 0.126m

Therefore, the distance the spring is stretched from its unstretched length (which is the same as the extension of the spring) is 0.126m

3 0
3 years ago
Pls help my science teacher is a horrible teacher...
Anettt [7]

First of all, looks like your teacher is indeed pretty horrible. Secondly, the constraints to consider would be proper weight distribution, methods to minimize excessive motion of the building structure, and quantities such as volume and density, which would help in determining the optimal structure. Keeping the frequency of oscillation for a building low in case of an earthquake or natural disaster would also be a priority.

4 0
3 years ago
In vacuum , the shorter the wavelength of an electromagnetic wave is , the:
skad [1K]

Answer:

Higher its Energy

Explanation:

7 0
2 years ago
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