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gtnhenbr [62]
3 years ago
15

Conductors of large instrumental ensembles use thin stick called a __________ to help performers keep time.

Physics
2 answers:
Dmitry [639]3 years ago
6 0
I think the word is "baton"
kkurt [141]3 years ago
3 0
Your answer is:

baton

Hope this helps have a great day =D
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A record of travel along a straight path is as follows: 1. Start from rest with constant acceleration of 2.65 m/s2 for 17.0 s. 2
nlexa [21]
Hello

Let's solve the problem in the three different steps

1) Uniformly accelerated motion, with acceleration a_1 = 2.65~m/s^2 and for a total time of t_1=17~s. The body is initially at rest, so the distance covered is given by
S= \frac{1}{2}a_1t_1^2=382.9~m
Calling v_f and v_i the final and initial velocity, and since the v_i=0~m/s because the body starts from rest, we can use
a= \frac{v_f-v_i}{t}
to find the final velocity after this first leg:
v_{f}=v_i+a_1t_1=45~m/s
And the average velocity in this first leg is
v_1= \frac{v_f+v_i}{2}=22.5~m/s

2) Uniform motion. The velocity is constant and it is equal to the final velocity of the first leg: v_2=45~m/s. This is also the average velocity of the second leg. 
The total time of this second leg is t_2=1.60~min = 96~s. The distance covered is given by
S_2=v_2t_2=45~m/s \cdot 96~s=4320~m

3) Uniformly decelerated motion, with constant deceleration of a_3=-9.39~m/s^2 and for a total time of t_3=4.8~s. Here, the initial velocity of the body is the final velocity of the previous leg, i.e. v_i=45~m/s. Therefore, the distance covered in this leg is given by
S_3=v_i t_3 + \frac{1}{2} a_3 t^2 =107.8~m
The final velocity in this leg is given by
v_f=v_i+at=45~m/s-9.39~m/s^2 \cdot 4.8~s = -0.07~m/s
The negative sign means that after decelerating, the body has started to go in the opposite direction. Similarly to step 1, the average velocity in this leg is given by
v_3 =  \frac{1}{2}(v_f+v_i)=  \frac{1}{2}(-0.07~m/s+45~m/s)=  22.5~m/s

4) Finally, the total distance covered in the motion is
S=S_1+S_2+S_3=382.9~m+4320~m+107.8~m=4810.7~m
To find the average velocity, we must "weigh" the average velocity of each leg for the correspondent time of that leg:
v_{ave}= \frac{v_1t_1+v_2t_2+v_3t_3}{t_1+t_2+t_3}=40.8~m/s
8 0
3 years ago
A flashlight bulb is connected to a dry cell of voltage 2.25 V. It draws 35.0 mA (1000 mA = 1 A). What is its resistance?
Anarel [89]

The answer & explanation for this question is given in the attachment below.

6 0
3 years ago
Two balls are kicked with the same initial speeds. Ball A was kicked at the angle 20° above horizontal and ball B was kicked at
GREYUIT [131]

Answer:

Ball A

Explanation:

Let the initial speed of the balls be u .

Angle of projection for ball A = 20°

Angle of projection for ball B = 75°

As we know that at highest point, the ball has only horizontal speed which always remains constant throughout the motion because the acceleration in horizontal direction is zero.

Speed of ball A at highest point = u Cos 20° = 0.94 u

Speed of ball B at highest point = u Cos 75° = 0.26 u

So, the ball A has bigger speed than B.

5 0
3 years ago
When two positive charges are brought close together, what happens to the
telo118 [61]

Answer:

they will move away from each other

7 0
3 years ago
A car is strapped to a rocket (combined mass = 661 kg), and its kinetic energy is 66,120 J.
labwork [276]

Answer:

9.4 m/s

Explanation:

According to the work-energy theorem, the work done by external forces on a system is equal to the change in kinetic energy of the system.

Therefore we can write:

W=K_f -K_i

where in this case:

W = -36,733 J is the work done by the parachute (negative because it is opposite to the motion)

K_i = 66,120 J is the initial kinetic energy of the car

K_f is the final kinetic energy

Solving,

K_f = K_i + W=66,120+(-36,733)=29387 J

The final kinetic energy of the car can be written as

K_f = \frac{1}{2}mv^2

where

m = 661 kg is its mass

v is its final speed

Solving for v,

v=\sqrt{\frac{2K_f}{m}}=\sqrt{\frac{2(29,387)}{661}}=9.4 m/s

4 0
3 years ago
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