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VARVARA [1.3K]
3 years ago
10

Simplify the expression [9+2(2-4)]/3

Physics
1 answer:
il63 [147K]3 years ago
4 0

Answer:

5/3

Explanation:

[9+2(2-4)]/3

BODMAS

[9+2(-2)]/3

[9-4]/3

5/3

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A proton moving eastward with a velocity of 5. 0 × 103 m/s enters a magnetic field of 0. 20 t pointing northward. what is the ma
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The magnitude of the force is 1.6*10^-16 N. The direction of force is upward.

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6 0
2 years ago
3. A sprinter leaves the starting blocks with an acceleration of 4.5 m/s2. What is the
UkoKoshka [18]

Hi there! :)

\large\boxed{v_{f} = 18 m/s}

Use the following kinematic equation to solve for the final velocity:

v_{f} = v_{i} + at

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v_{f} = at

Plug in the given acceleration and time:

v_{f} = 4.5 * 4 = 18 m/s

5 0
3 years ago
Why do astronomers use the word on to describe angles on the sky rather than angles in the sky?
Sever21 [200]
<span>The use of the word on instead of the word in when referring to the angular distance between celestial objects comes about because all of the objects appear to be on the celestial sphere and at an indeterminable distance. While we know that objects are at different distances in the sky, their distance from Earth is irrelevant in determining the angular distance between the two objects as viewed from Earth.</span>
8 0
3 years ago
If a player through a basketball to the target with an initial velocity of 17 m/s making an angle of 30 degrees with the horizon
Svetllana [295]

Answer:

The final position made with the vertical is 2.77 m.

Explanation:

Given;

initial velocity of the ball, V = 17 m/s

angle of projection, θ = 30⁰

time of motion, t = 1.3 s

The vertical component of the velocity is calculated as;

V_y = Vsin \theta\\\\V_y = 17 \times sin(30)\\\\V_y = 8.5 \ m/s

The final position made with the vertical (Yf) after 1.3 seconds is calculated as;

Y_f = V_yt  - \frac{1}{2}g t^2\\\\Y_f = (8.5 \times 1.3 ) - (\frac{1}{2} \times 9.8 \times 1.3^2)\\\\Y_f = 11.05 \ - \ 8.281\\\\Y_f = 2.77 \ m

Therefore, the final position made with the vertical is 2.77 m.

3 0
3 years ago
Which type of waves from the electromagnetic spectrum are used in heat lamps and heat sensing devices?
Alex73 [517]
Infrared waves are used in heat lamps and other heat sensing devices. Infrared waves or commonly known as Infrared radiations (IR) is the type of electromagnetic radiation we encounter most in our everyday life. Heat lamps are electrical devices which emit infrared radiation.
7 0
3 years ago
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