All categories

3 years ago

What are some examples of aerobic exercise?

Physics

1 answer:

Firdavs [7]3 years ago

4 0

Answer:

swimming, cycling, and jogging

Explanation:

You might be interested in

A spring-loaded gun shoots a plastic ball with a launch speed of 2.0 m/s. If the spring is replaced with a new spring having twi

sattari [20]

244Explanation:

ttyyyyhhvkdfgjdjj

3 0

4 years ago

What is the value of x?<br><br> 6x−2(x 4)=12

Ymorist [56]

Answer: The value of x is -6.

Explanation:

To calculate the value of 'x', we need to solve each function happening inthe equation.

The equation provided to us is $6x-2(4x)=12$

To solve this, we will multiply 4x with 2 and then subtract the like terms and finally, we evaluate the value of 'x'.

$6x-8x=12\\\\-2x=12\\\\x=\frac{12}{-2}\\\\x=-6$

Hence, the value of x will be -6.

5 0

4 years ago

Read 2 more answers

Please help Draw where the rays will go and label the type of mirror.

lozanna [386]

Answer:

Explanation:

Please check the picture and consider straight lines

8 0

4 years ago

Based on the graph of velocity over time, which could be the initial velocity and the final velocity for this graph?

dolphi86 [110]

The graph is showing a constant velocity as the line is horizontal. The initial speed is equal to the final speed as there is no changing in acceleration. The second statement is the correct statement

Initial = 2.5 m/s
Final = 2.5 m/s

3 0

3 years ago

Read 2 more answers

The earth rotates once per day about an axis passing through the north and south poles, an axis that is perpendicular to the pla

Vladimir [108]

Answer

given,

Radius of sphere = 6.38 × 10⁶ m

time = 1 day = 86400 s

$\omega = \dfrac{2\pi}{T}$

$\omega = \dfrac{2\pi}{ 86400 }$

$\omega = 7.272 \times 10^{-5}\ rad/s$

a) at equator

$v = R_E \omega$

$v = 6.38 \times 10^6\times 7.272 \times 10^{-5}$

v = 464 m/s

acceleration of the person

$a = \omega^2R_E$

$a = (7.272 \times 10^{-5})^2(6.38 \times 10^6)$

$a = 0.03374 m/s^2$

b) at a latitude of 61.0 ° north of the equator.

$R = R_E cos \theta$

$R = 6.38 \times 10^6\times cos 61^0$

$R = 3.093 \times 10^6 m$

$v = R \omega$

$v = 3.093 \times 10^6 \times 7.272 \times 10^{-5}$

$v = 225 m/s$

acceleration of the person

$a = \omega^2R_E$

$a = (7.272 \times 10^{-5})^2(3.093 \times 10^6 )$

$a = 0.01635 m/s^2$

4 0

3 years ago