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Which of the following is NOT an exercise myth?

galben [10]

I believe it would be A, (B) just because a women trains doesn't mean she'll bulk, the may become thinner, or more tone. No to C because You have to change up and make workouts more difficult to help improve your body, and no to D because a workout doesn't have to be strenuous, you can be careful and not crazy about working out and still help your body. hope this helped

8 0

3 years ago

When a wave has λ=3 m and f=15 Hz, what is the speed of the wave?

Tamiku [17]

Wave speed = (wavelength) x (frequency)

Wave speed = (3 m) x (15 Hz)

<em>Wave speed = 45 m/s</em>

5 0

3 years ago

Bullets from two revolvers are fired with the same velocity. The bullet from gun #1 is twice as heavy as the bullet from gun #2.

IrinaVladis [17]

Answer:

option C

Explanation:

Let mass of the bullet be m and velocity be v

mass of gun be M and bullet be V

now,

using conservation of momentum for gun 1

(M+m) V' = 2 mv + 3 MV

V' = 0

3 M V = - 2 mv

momentum of gun 1 =- 2 mv---------(1)

now for gun 2

(M+m) V' = mv + MV

V' = 0

M V = - mv

momentum of gun 1 = -mv-----------(2)

dividing equation (1) by (2)

$\dfrac{P_m1}{P_m2} = \dfrac{- 2mv}{-mv}$

$\dfrac{P_m1}{P_m2} = \dfrac{2}{1}$

the correct answer is option C

7 0

3 years ago

The flywheel of a steam engine runs with a constant angular velocity of 190 rev/min. When steam is shut off, the friction of the

Ivanshal [37]

Answer:

(a) $\alpha = - 1.32\ rev/m^{2}$

(b) $\theta = 13674\ rev$

(c) $\alpha_{tan} = 8.75\times 10^{- 4}\ m/s^{2}$

(d) $a = 22.458\ m/s^{2}$

Solution:

As per the question:

Angular velocity, $\omega = 190\ rev/min$

Time taken by the wheel to stop, t = 2.4 h = $2.4\times 60 = 144\ min$

Distance from the axis, R = 38 cm = 0.38 m

Now,

(a) To calculate the constant angular velocity, suing Kinematic eqn for rotational motion:

$\omega' = \omega + \alpha t$

$omega'$ = final angular velocity

$\omega$ = initial angular velocity

$\alpha$ = angular acceleration

Now,

$0 = 190 + \alpha \times 144$

$\alpha = - 1.32\ rev/m^{2}$

Now,

(b) The no. of revolutions is given by:

$\omega'^{2} = \omega^{2} + 2\alpha \theta$

$0 = 190^{2} + 2\times (- 1.32) \theta$

$\theta = 13674\ rev$

(c) Tangential component does not depend on instantaneous angular velocity but depends on radius and angular acceleration:

$\alpha_{tan} = 0.38\times 1.32\times \frac{2\pi}{3600} = 8.75\times 10^{- 4}\ m/s^{2}$

(d) The radial acceleration is given by:

$\alpha_{R} = R\omega^{2} = 0.32(80\times \frac{2\pi}{60})^{2} = 22.45\ rad/s$

Linear acceleration is given by:

$a = \sqrt{\alpha_{R}^{2} + \alpha_{tan}^{2}}$

$a = \sqrt{22.45^{2} + (8.75\times 10^{- 4})^{2}} = 22.458\ m/s^{2}$

5 0

3 years ago

Part e a small toy cart equipped with a spring bumper rolls toward a wall with a speed of v. the cart rebounds from the wall, wi

jeka94

Answer:

Δp = -2 p₀

Explanation:

The momentum is defined by

p = m v

In this case we write the initial and final momentum, we take as positive the direction towards the wall.

p₀ = m v

p_f = m (-v)

the negative sign is because the car is bouncing off the wall

the change of the moment is

Δp = p_f - p₀

Δp = - m v - m v

Δp = -2 mv

Δp = -2 p₀

we see that the change of moment is twice the moment, in the attachment we can see the vectors of these changes and the sign indicates the direction of the change at the moment

5 0

3 years ago