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2 years ago

8

Does gender affect your aerobic capacity?

1 answer:

alekssr [168]2 years ago

8 0

To a degree yes, the mass and volume of a male is far greater than that of the opposite gender. As well as the type of aerobics , but with practice and daily aerobics activities one can become very flexible

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A person gives a shopping cart an initial push along a horizontal floor to get it moving, and then lets go. The cart travels for

mash [69]

Answer:

Only a backward force is acting, no forward force.

Explanation:

Once released from the initial push, in absence of friction, the shopping cart would continue moving forward at a constant speed forever.
As it would move at a constant speed, no net force would be acting on it.
So, if it is gradually slowing, there must be a net force producing an acceleration in a direction opposite to the movement.
This force is the kinetic friction force, and is the only force acting on the cart in the horizontal direction.
As any friction force, opposes to the relative movement between the cart and the horizontal floor, which means that is directed backward.
This is consistent with the direction of the acceleration of the cart.

8 0

3 years ago

Which of the following is the FINAL step in a forecasting system?

Lelechka [254]

Answer:

gather the data needed to make the forecast

3 0

2 years ago

A 0.106-A current is charging a capacitor that has square plates 6.00 cm on each side. The plate separation is 4.00 mm. (a) Find

FrozenT [24]

Answer:

The time rate of change of flux is $1.34 \times 10^{10}$ $\frac{V}{s}$

Explanation:

Given :

Current $I = 0.106$ A

Area of plate $A = 36 \times 10^{-4}$ $m^{2}$

Plate separation $d = 4 \times 10^{-3}$ m

(A)

First find the capacitance of capacitor,

$C = \frac{\epsilon _{o} A }{d}$

Where $\epsilon _{o} = 8.85 \times 10^{-12}$

$C = \frac{8.85 \times 10^{-12 } \times 36 \times 10^{-4} }{4 \times 10^{-3} }$

$C = 7.9 \times 10^{-12}$ F

But $C = \frac{Q}{V}$

Where $Q = It$

$C = \frac{It}{V}$

$V = \frac{It}{C}$

Now differentiate above equation wrt. time,

$\frac{dV}{dt} = \frac{I}{C}$

$= \frac{0.106}{7.9 \times 10^{-12} }$

$= 1.34 \times 10^{10}$ $\frac{V}{s}$

Therefore, the time rate of change of flux is $1.34 \times 10^{10}$ $\frac{V}{s}$

8 0

2 years ago

A train slows down as it rounds a sharp horizontal turn, going from 94.0 km/h to 46.0 km/h in the 17.0 s that it takes to round

Svetllana [295]

Answer:

1.41 m/s^2

Explanation:

First of all, let's convert the two speeds from km/h to m/s:

$u = 94.0 km/h \cdot \frac{1000 m/km}{3600 s/h} = 26.1 m/s$

$v=46.0 km/h \cdot \frac{1000 m/km}{3600 s/h}=12.8 m/s$

Now we find the centripetal acceleration which is given by

$a_c=\frac{v^2}{r}$

where

v = 12.8 m/s is the speed

r = 140 m is the radius of the curve

Substituting values, we find

$a_c=\frac{(12.8 m/s)^2}{140 m}=1.17 m/s^2$

we also have a tangential acceleration, which is given by

$a_t = \frac{v-u}{t}$

where

t = 17.0 s

Substituting values,

$a_t=\frac{12.8 m/s-26.1 m/s}{17.0 s}=-0.78 m/s^2$

The two components of the acceleration are perpendicular to each other, so we can find the resultant acceleration by using Pythagorean theorem:

$a=\sqrt{a_c^2+a_t^2}=\sqrt{(1.17 m/s^2)+(-0.78 m/s^2)}=1.41 m/s^2$

6 0

2 years ago

Read 2 more answers

How much force is needed to accelerate a vehicle with a mass of 1000 kg at a rate<br> of 5 m/s2?

Mashcka [7]

The force needed to accelerate a vehicle with a mass of 1000kg at a rate of 5m/s2 would be 5000

8 0

2 years ago