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ch4aika [34]
3 years ago
7

Want brainlyist whats number am i thinking of from 1 to 100

Physics
2 answers:
tatuchka [14]3 years ago
4 0

Answer:

50 out of 100 (ik its right)

BlackZzzverrR [31]3 years ago
3 0

Answer:

57? 17? 7? 77? 37? 47? 87? 97? 27? 67?

Explanation:

id'k

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A negative test charge experiences a force to the right as a result of an electric field. Which is the best conclusion to draw
Alona [7]

Answer:a

Explanation:

4 0
3 years ago
3. A block of mass m1=1.5 kg on an inclined plane of an angle of 12° is connected by a cord over a mass-less, frictionless pulle
Lena [83]

Answer:

\mu=0.377

Explanation:

we need to start by drawing the free body diagram for each of the masses in the system. Please see attached image for reference.

We have identified in green the forces on the blocks due to acceleration of gravity (w_1 and  w_2) which equal the product of the block's mass times "g".

On the second block (m_2), there are just two forces acting: the block's weight  (m_2\,*\,g) and the tension (T) of the string. We know that this block is being accelerated since it has fallen 0.92 m in 1.23 seconds. We can find its acceleration with this information, and then use it to find the value of the string's tension (T). We would need both these values to set the systems of equations for block 1 in order to find the requested coefficient of friction.

To find the acceleration of block 2 (which by the way is the same acceleration that block 1 has since the string doesn't stretch) we use kinematics of an accelerated object, making use of the info on distance it fell (0.92 m) in the given time (1.23 s):

x_f-x_i=v_i\,t-\frac{1}{2} a\,t^2 and assume there was no initial velocity imparted to the block:

x_f-x_i=v_i\,t-\frac{1}{2} a\,t^2\\-0.92\,m=0\,-\frac{1}{2} a\,(1.23)^2\\a=\frac{0.92\,*\,2}{1.23^2} \\a=1.216 \,\frac{m}{s^2}

Now we use Newton's second law in block 2, stating that the net force in the block equals the block's mass times its acceleration:

F_{net}=m_2\,a\\w_2-T=m_2\,a\\m_2\,g-T=m_2\,a\\m_2\,g-m_2\,a=T\\m_2\,(g-a)=T\\1.2\,(9.8-1.216)\,N=T\\T=10.3008\,N

We can round this tension (T) value to 10.3 N to make our calculations easier.

Now, with the info obtained with block 2 (a - 1.216 \frac{m}{s^2}, and T = 10.3 N), we can set Newton's second law equations for block 1.

To make our study easier, we study forces in a coordinate system with the x-axis parallel to the inclined plane, and the y-axis perpendicular to it. This way, the motion in the y axis is driven by the y-component of mass' 1 weight (weight1 times cos(12) -represented with a thin grey trace in the image) and the normal force (n picture in blue in the image) exerted by the plane on the block. We know there is no acceleration or movement of the block in this direction (the normal and the x-component of the weight cancel each other out), so we can determine the value of the normal force (n):

n-m_1\,g\,cos(12^o)=0\\n=m_1\,g\,cos(12^o)\\n=1.5\,*\,9.8\,cos(12^o)\\n=14.38\,N

Now we can set the more complex Newton's second law for the net force acting on the x-axis for this block. Pointing towards the pulley (direction of the resultant acceleration a), we have the string's tension (T). Pointing in the opposite direction we have two forces: the force of friction (<em>f</em> ) with the plane, and the x-axis component of the block's weight (weight1 times sin(12)):

F_{net}=m_1\,a\\T-f-w_1\,sin(12)=m_1\,a\\T-w_1\,sin(12)-m_1\,a=f\\f=[10.3-1.5\,*\,9.8\,sin(12)-1.5\,*1.216]\,N\\f=5.42\,N

And now, we recall that the force of friction equals the product of the coefficient of friction (our unknown \mu) times the magnitude of the normal force (14.38 N):

f=\mu\,n\\5.42\,N=\mu\,*\,14.38\,N\\\mu=\frac{5.42}{14.38}\\\mu=0.377

with no units.

4 0
3 years ago
A 45.0 kg student runs at a constant velocity up the incline 6 metres high. If the power output of the student is 1.50 x 103 W,
serg [7]

Answer:

The time taken is 1.76 s.

Explanation:

mass of student, m = 45 kg

height, h = 6 m

power, P = 1500 W

Power is defined as the rate of doing work.

Work done in moving height is

W = m g h

Let the time is t.

P = m g h / t

1500 = 45 x 9.8 x 6 / t

t = 1.76 s

3 0
3 years ago
The velocity of a particle moving along the xaxis varies according to the expres-sion v = 40 - 5t², where vis in meters per seco
Afina-wow [57]

(A) -10m/s^2

(B) the average acceleration in the specified time in interval Δt = tb - ta =2.0sis

The rate at which the velocity changes is referred to as average acceleration. To determine the average acceleration of something, we divide the change in velocity by the amount of time that has passed.

For instance, the average acceleration of a crazy ball would be 20 cm/s/s if its velocity increased from 0 to 60 cm/s in 3 seconds. The marble's velocity will therefore rise by 20 cm/s per second as a result.

As a result, we can see that the definition of average acceleration is the ratio of change in velocity to change in time for a certain interval. The average acceleration is calculated over a given period of time and contrasted with instantaneous acceleration. The final velocity less the starting velocity divided by the total time is the average acceleration.

To learn more about average acceleration please visit-
brainly.com/question/17352997
#SPJ9

4 0
1 year ago
Which aspect of a wave must remain constant as long as the medium of the wave doesn’t change? A. Frequency B. Amplitude C. Veloc
scZoUnD [109]
A wave is a disturbance that moves along a medium from one end to the other. If one watches an ocean wave moving along the medium (the ocean water), one can observe that the crest of the wave is moving from one location to another over a given interval of time. The crest is observed to cover distance. The speed of an object refers to how fast an object is moving and is usually expressed as the distance traveled per time of travel. In the case of a wave, the speed is the distance traveled by a given point on the wave (such as a crest) in a given interval of time. In equation form,
7 0
3 years ago
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