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4 years ago

11

Calculate the average velocity of a dancer who moves 5 m toward the left of the stage over the course of 15 s. ** Velocity = dis

placement/time Question 1 options: A. 3 m/s B. 1/3 m/s C. 1/3 m/s west D. 3 m/s west

1 answer:

Alex73 [517]4 years ago

3 0

Answer:

B

Explanation:

Velocity=disp/time

V=5m/15s

V=1/3 m/s

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Based on the graph what kind of relationship exists between the change in temperature of the water in the gravitational potentia

Musya8 [376]

Answer: It is a direct relationship

Explanation: I just did that question.

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4 years ago

Read 2 more answers

A body moving with an initial velocity of 30m/s accelerates uniformly at the rate of 10m/s . what is the distance covered during

nikdorinn [45]

Answer:

The distance covered by the body is, S = 800 m

Explanation:

Given data,

The initial velocity of the body, u = 30 m/s

The acceleration of the body, a = 10 m/s²

Let the time period of travel be, t = 10 s

Using the II equations of motion,

S = ut + ½ at²

Substituting the given values,

S = 30 x 10 + ½ x 10 x 10²

S = 800 m

Hence, the distance covered by the body is, S = 800 m

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3 years ago

A force platform is a tool used to analyze the performance of athletes measuring the vertical force that the athlete exerts on t

Phoenix [80]

Answer:

a. $I=981.34 N*s$

b. $v_f=3.96 m/s$

c. $v_{f1}=3.63m/s$

d. $y_f=0.673m$

Explanation:

Given: $m=67kg$ , $h=0.720m$ , $0$

a.

$I=\int\limits^{t_1}_{t_2} {F(t)} \, dt$

$F(t)=9200*t-11500t^2$

$I=\int\limits^{0.8s}_{0s}{9200*t-11500*t^2} \, dt$

$I=4600*t^2-3833.3*t^3|(0.80,0)$

$I=2944-1962.66=981.35$

$I=981.34 N*s$

b.

$v_f^2=v_i^2+a*y'$

Starting from the rest

$v_f^2=0+2*9.8m/s^2*0.80s$

$v_f^2=15.68$

$v_f=\sqrt{15.68m^2/s^2}=3.96 m/s$

c.

$I_{total}=p_f$

$I_1-m*g*d=m*v_{f1}-m*v_f$

$981.34-67kg*9.8m/s^2*0.720=67.0kg*v_{f1}-67.0kg*(-3.96m/s)$

Solve to vf

$v_{f1}=3.63m/s$

d.

$v_f^2=v_i^2+2*a*y_f'$

$y_f'=v_i/2*a =(3.63m/s)^2/2*9.8m/s^2$

$y_f=0.673m$

7 0

3 years ago

Find the magnitude of u × v and the unit vector parrallel to u × v in the direction of u × v. u = 2i + 2j - k, v = -i + k

IrinaVladis [17]

Answer:

magnitude = 3

unit vector = $\frac{2i}{3} - \frac{j}{3} - \frac{2k}{3}$

Explanation:

Given vectors:

u = 2i + 2j - k

v = -i + k = -i + 0j + k

(a) u x v is the cross product of u and v, and is given by;

$u X v = \left[\begin{array}{ccc}i&j&k\\2&2&-1\\-1&0&1\end{array}\right]$

u x v = i(2+0) - j(2 - 1) + k(0 - 2)

u x v = 2i - j - 2k

Now the magnitude of u x v is calculated as follows:

| u x v | = $\sqrt{2^2 + (-1)^2 + (-2)^2}$

| u x v | = $\sqrt{4 + 1 + 4}$

| u x v | = $\sqrt{9}$

| u x v | = 3

Therefore, the magnitude of u x v is 3

(b) The unit vector û parallel to u x v in the direction of u x v is given by the ratio of u x v and the magnitude of u x v. i.e

û = $\frac{u X v}{|u X v|}$

u x v = 2i - j - 2k [<em>calculated in (a) above</em>]

|u x v| = 3 [<em>calculated in (a) above</em>]

∴ û = $\frac{2i - j - 2k}{3}$

∴ û = $\frac{2i}{3} - \frac{j}{3} - \frac{2k}{3}$

4 0

3 years ago

An ideal horizontal spring-mass system has a mass of 1.0 kg and a spring with constant 78 N/m. It oscillates with a period of 0.

steposvetlana [31]

Answer:

T = 0.71 seconds

Explanation:

Given data:

mass m = 1Kg, spring constant K = 78 N/m, time period of oscillation T = 0.71 seconds.

We have to calculate time period when this same spring-mass system oscillates vertically.

As we know

$T = 2\pi \sqrt{\frac{m}{K} }$

This relation of time period is true under every orientation of the spring-mass system, whether horizontal, vertical, angled or inclined. Therefore, time period of the same spring-mass system oscillating vertically too remains the same.

Therefore, T = 0.71 seconds

6 0

3 years ago