Kate is testing whether caffeine affect your heart rate independent variable or dependent variable

Let’s tie this all together. It makes sense that, if the rope force remains greater than the gravitational force, the child keep

Pavlova-9 [17]

Answer:

If child weight is equal to rope force then child will move with uniform speed

or we can say that the child will remain at rest in his position

Explanation:

As we know that child is hanging by rope

so here there will be two forces on the child

1) Weight or gravitational force which act vertically downwards

2) Tension in the rope which act vertically upwards

Now if child will accelerate upwards then tension force must be more than the weight of the child

If tension force is less than the weight then child will decelerate and his speed will decrease

if tension force is equal to child weight then in that case the child will remain at rest or it will move with same speed

8 0

3 years ago

A student evaluates a weight loss program by calculating the number of times he would need to climb a 14.0 m high flight of step

Liula [17]

Answer:

400 trips

Explanation:

Mechanical energy needed to climb 14 m by a man of 68 kg

= mgh

= 68 x 9.8 x 14

= 9330 J

1 Kg of fat releases 3.77 x 10⁷ J of energy

.45 kg of fat releases 1.6965 x 10⁷ J of energy

22% is converted into mechanical energy

so 22% of 1.6965 x 10⁷ J

= 3732.3 x 10³ J of mechanical energy will be available for mechanical work.

one trip of climbing of 14 m requires 9330 J of mechanical energy

no of such trip possible with given mechanical energy

= 3732.3 x 10³ / 9330

= 400 trips

7 0

3 years ago

You are driving on the highway, and you come to a steep downhill section. As you roll down the hill, you take your foot off the

Vinvika [58]

Answer:

Weight of the car, normal force, drag force

Explanation:

The forces acting on the car are:

The normal force which acts perpendicularly to the downhill plane
The weight of the car which acts vertically downwards
The drag force due to air resistance which acts in opposition to the motion of the car

Friction is ignored, so the force due to friction is assumed negligible

6 0

3 years ago

Two black holes (the remains of exploded stars), separated by a distance of

jolli1 [7]

The largest mass is 4.7 x 10³⁰ kg and the smallest mass is 5 x 10²⁹ kg.

The given parameters;

distance between the two black holes, r = 10 AU = 1.5 x 10¹² m
gravitational force between the two black holes, F = 6.9 x 10²⁵ N.
combined mass of the two black holes = 5.20 x 10³⁰ kg

The product of the two masses is calculated from Newton's law of universal gravitational as follows;

$F = \frac{Gm_1m_2}{r^2} \\\\m_1m_2 = \frac{Fr^2}{G} \\\\m_1m_2 = \frac{(6.9\times 10^{25}) \times (1.5\times 10^{12})^2}{6.67\times 10^{-11}} \\\\m_1m_2 = 2.328 \times 10^{60} \ kg^2$

The sum of the two masses is given as;

m₁ + m₂ = 5.2 x 10³⁰ kg

m₂ = 5.2 x 10³⁰ kg - m₁

The first mass is calculated as follows;

m₁(5.2 x 10³⁰ - m₁) = 2.328 x 10⁶⁰

5.2 x 10³⁰m₁ - m₁² = 2.328 x 10⁶⁰

m₁² - 5.2 x 10³⁰m₁ + 2.328 x 10⁶⁰ = 0

solve the quadratic equation using formula method;

a = 1, b =- 5.2 x 10³⁰, c = 2.328 x 10⁶⁰

$m_1 = \frac{-b \ \ +/- \ \ \sqrt{b^2 - 4ac} }{2a} \\\\m_1 = \frac{-(-5.2\times 10^{20}) \ \ +/- \ \ \sqrt{(-5.2\times 10^{20})^2 - 4(1\times 2.328\times 10^{60})} }{2(1)} \\\\m_1 = 4.7 \times 10^{30} \ kg \ \ or \ \ 4.9 \times 10^{29} \ kg$