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4 years ago

7

You decide to ride your bike for 30 minutes, raising your heart rate to the middle of your target heart rate zone. What type of

activity is this an example of?

2 answers:

a_sh-v [17]4 years ago

4 0

It would have to be cardio i think not totally sure

GaryK [48]4 years ago

4 0

Answer:

Cardio aerobic

Explanation:

Raising the heart rate uses a lot of oxygen and since it is increasing the heart rate, that means the heart is pumping blood at a faster rate than normal. Thus it is a cardio aerobic activity, where the rate is increased that burns a lot of oxygen . It also increases the metabolic system of the body and helps to maintain a healthy body.

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A physics student stands on a cliff overlooking a lake and decides to throw a softball to her friends in the water below. She th

aliina [53]

Answer:

58.5 m

Explanation:

First of all, we need to find the total time the ball takes to reach the water. This can be done by looking at the vertical motion only.

The initial vertical velocity of the ball is

$u_y = u sin \theta$

where

u = 21.5 m/s is the initial speed

$\theta=33.5^{\circ}$ is the angle

Substituting,

$u_y = (21.5) sin 33.5^{\circ} =11.9 m/s$

The vertical position of the ball at time t is given by

$y = h + u_y t + \frac{1}{2}gt^2$

where

h = 13.5 m is the initial heigth

$g = -9.8 m/s^2$ is the acceleration of gravity (negative sign because it points downward)

The ball reaches the water when y = 0, so

$0 = h + u_yt +\frac{1}{2}gt^2\\0 = 13.5 +11.9 t - 4.9t^2$

Which gives two solutions: t = 3.27 s and t = -0.84 s. We discard the negative solution since it is meaningless.

The horizontal velocity of the ball is

$u_y = u cos \theta = (21.5) cos 33.5^{\circ} =17.9 m/s$

And since the motion along the horizontal direction is a uniform motion, we can find the horizontal distance travelled by the ball as follows:

$d= u_x t = (17.9)(3.27)=58.5 m$

3 0

3 years ago

State Pascal's principle of pressure . please help due tomorrow

Romashka [77]

Answer:

Pascal's law says that pressure applied to an enclosed fluid will be transmitted without a change in magnitude to every point of the fluid and to the walls of the container.

Explanation:

The pressure at any point in the fluid is equal in all directions.

3 0

3 years ago

A procedure called______is used to find the volume of a rock

KengaRu [80]

Displacement is usually how the messure the rock

8 0

3 years ago

Read 2 more answers

An object has the acceleration graph shown in (Figure 1). Its velocity at t=0s is vx=2.0m/s. Draw the object's velocity graph fo

timama [110]

Answer:

Explanation:

We may notice that change in velocity can be obtained by calculating areas between acceleration lines and horizontal axis ("Time"). Mathematically, we know that:

$v_{b}-v_{a} = \int\limits^{t_{b}}_{t_{a}} {a(t)} \, dt$

$v_{b} = v_{a}+ \int\limits^{t_{b}}_{t_{a}} {a(t)} \, dt$

Where:

$v_{a}$ , $v_{b}$ - Initial and final velocities, measured in meters per second.

$t_{a}$ , $t_{b}$ - Initial and final times, measured in seconds.

$a(t)$ - Acceleration, measured in meters per square second.

Acceleration is the slope of velocity, as we know that each line is an horizontal one, then, velocity curves are lines with slopes different of zero. There are three region where velocities should be found:

Region I (t = 0 s to t = 4 s)

$v_{4} = 2\,\frac{m}{s} +\int\limits^{4\,s}_{0\,s} {\left(-2\,\frac{m}{s^{2}} \right)} \, dt$

$v_{4} = 2\,\frac{m}{s}+\left(-2\,\frac{m}{s^{2}} \right) \cdot (4\,s-0\,s)$

$v_{4} = -6\,\frac{m}{s}$

Region II (t = 4 s to t = 6 s)

$v_{6} = -6\,\frac{m}{s} +\int\limits^{6\,s}_{4\,s} {\left(1\,\frac{m}{s^{2}} \right)} \, dt$

$v_{6} = -6\,\frac{m}{s}+\left(1\,\frac{m}{s^{2}} \right) \cdot (6\,s-4\,s)$

$v_{6} = -4\,\frac{m}{s}$

Region III (t = 6 s to t = 10 s)

$v_{10} = -4\,\frac{m}{s} +\int\limits^{10\,s}_{6\,s} {\left(2\,\frac{m}{s^{2}} \right)} \, dt$

$v_{10} = -4\,\frac{m}{s}+\left(2\,\frac{m}{s^{2}} \right) \cdot (10\,s-6\,s)$

$v_{10} = 4\,\frac{m}{s}$

Finally, we draw the object's velocity graph as follows. Graphic is attached below.

3 0

4 years ago

A proton is projected into a magnetic field that is directed along the positive x axis. Find the direction of the magnetic force

bulgar [2K]

Te direction of the magnetic force for the velocity of the proton in the

-ve y direction will be +ve z direction.

As we know that the right-hand rule is based on the relation of magnetic fields and the forces that they exert on moving charges.When a charged particle moves under a magnetic field, it exerts a force on the particle, which is not in the same direction but different than the direction of the magnetic field.Under the right-hand rule, if we point our pointer finger in the direction of the charged particle is moving and the middle finger is representing the direction of the magnetic field then our thumb depicts the direction of the magnetic force which is exerted on the charged particle.

So, we are given that the direction of the velocity of the proton is in the negative y direction and the direction of the magnetic field is in the positive x direction, so the magnetic force is acting in the positive z direction.

To know more about the right-hand rule refer to the link brainly.com/question/9750730?referrer=searchResults.

#SPJ4

4 0

2 years ago