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grandymaker [24]
3 years ago
15

A brick with a mass of 20 kg and a weight of 196 N is released on a frictionless plane at an angle of 30°. What is the accelerat

ion of the brick?
A. 4.9 m/s2
B. 5.6 m/s2
C. 8.5 m/s2
D. 9.8 m/s2
Physics
1 answer:
Veronika [31]3 years ago
6 0

Answer:4.9m/s^2

Basically answer A

Explanation:

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How high would a projectile go if it was launched from ground level with an initial speed of 26 m/s at an angle of 30 degrees ab
tigry1 [53]

Answer:

Vy = 26 m/s sin 30 = 13 m/s      vertical speed

t = Vy / a = 13 m/s / 9.80 m/s^2 = 1.33 sec     time to reach Vy = 0

H = Vy t + 1/2 g t^2

H = 13 m/s * 1.33 sec - 1.33^2 * 9.8 / 2 m = 8.62 m

4 0
1 year ago
A reconnaissance plane flies 545 km away from
Nikolay [14]

Answer:

681.6/ms

Explanation:

A reconnaissance plane flies 545 km away from  its base at 568 m/s. then flies back to its base  at 852 m/s.

What is its average speed?

Answer in its of m/s

Avg speed of the round trip is

2*568*852/(568+852)= 681.6/ms

6 0
3 years ago
Considering the factors that affect gravitational pull, in which location would the gravitational pull from the earth be SMALLES
OleMash [197]
C. The higher the altitude the less gravity affects you
4 0
3 years ago
A student visits the beach and wants to explore how landforms, such as sand dunes, are the result of changes caused by wind. wha
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3 0
3 years ago
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A wheel has a radius of r = 2.0 m and it rolls down a smooth incline. The height of the incline is h = 8.0 m . What is the angul
zepelin [54]

The angular velocity of the wheel at the bottom of the incline is 4.429 rad/sec

The angular velocity (ω) of an object is the rate at which the object's angle position is changing in relation to time.

For a wheel attached to an incline angle, the angular velocity can be computed by considering the conservation of energy theorem.

As such the total kinetic energy (K.E) and rotational kinetic energy (R.K.E) at a point is equal to the total potential energy (P.E) at the other point.

i.e.

P.E = K.E + R.K.E

\mathbf{mgh = \dfrac{1}{2}m(r \times \omega)^2 + \dfrac{1}{2}\times I \times \omega^2}

\mathbf{gh = \dfrac{1}{2}(r \times \omega)^2 + \dfrac{1}{2}\times r^2 \times \omega^2}

\mathbf{2 \times \dfrac{gh}{r^2} =\omega^2 +  \omega^2}

\mathbf{2 \omega^2=2 \times \dfrac{9.81 \times 8 m }{2.0 ^2}  }

\mathbf{\omega^2=\dfrac{39.24 }{2}}

\mathbf{\omega=\sqrt{19.62 } \ rad/sec}

\mathbf{\omega=4.429 \ rad/sec}

Therefore, we can conclude that the angular velocity of the wheel at the bottom of the incline is 4.429 rad/sec

Learn more about angular velocity here:

brainly.com/question/1452612

4 0
2 years ago
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