What hormone does the pineal gland release

5. What is the velocity of a Nolan Catholic football player if he runs 200.0 m in 24.0 s?

Degger [83]

Answer:

$\boxed{ \bold{ \huge{ \boxed {\sf{8.33 \: m \:/s \: }}}}}$

Explanation:

Distance travelled = 200 metre

Time taken = 24 second

Velocity = ?

Finding the velocity 

$\boxed{ \sf{velocity = \frac{distance \: travelled \: }{time \: taken}}}$

$\dashrightarrow{ \sf{velocity = \frac{200 \: m}{24 \: s} }}$

$\dashrightarrow{ \sf{velocity = 8.33 \: m/s}}$

Hope I helped!

Best regards!

6 0

3 years ago

A man is standing on the edge of a 20.0 m high cliff. He throws a rock horizontally with an initial velocity of 10.0 m/s.

kherson [118]

Answer:

a. The rock takes 2.02 seconds to hit the ground

b. The rock lands at 20,2 m from the base of the cliff

Explanation:

Horizontal motion occurs when an object is thrown horizontally with an initial speed v from a height h above the ground. When it happens, the object moves through a curved path determined by gravity until it hits the ground.

The time taken by the object to hit the ground is calculated by:

$\displaystyle t=\sqrt{\frac{2h}{g}}$

The range is defined as the maximum horizontal distance traveled by the object and it can be calculated as follows:

$\displaystyle d=v.t$

The man is standing on the edge of the h=20 m cliff and throws a rock with a horizontal speed of v=10 m/s.

a,

The time taken by the rock to reach the ground is:

$\displaystyle t=\sqrt{\frac{2*20}{9.8}}$

$\displaystyle t=\sqrt{4.0816}$

t = 2.02 s

The rock takes 2.02 seconds to hit the ground

b.

The range is calculated now:

$\displaystyle d=10\cdot 2.02$

d = 20.2 m

The rock lands at 20,2 m from the base of the cliff

5 0

2 years ago

Derive an expression for the gravitational potential energy U(r) of the object-earth system as a function of the object's distan

Drupady [299]

Answer:

$U(r)=-\frac{Gm_Emr^2}{2R^3_E}$

Explanation:

We are given that

Gravitational force= $F_g=\frac{Gm_Emr}{R^3_E}$

r=0,U(0)=0

We know that

Gravitational potential energy= $-\int F_gdr$

$U(r)=-\int\frac{Gm_Emr}{R^3_E}dr$

$U(r)=-\frac{Gm_Em}{R^3_E}\times \frac{r^2}{2}+C$

Substitute r=0 ,U(0)=0

$0=0+C$

$C=0$

Substitute the value

$U(r)=-\frac{Gm_Emr^2}{2R^3_E}$

4 0

3 years ago

g A 1.45-kg block is pushed against a vertical wall by means of a spring (k = 860 N/m). The coefficient of static friction betwe

olga_2 [115]

Answer:

The minimum compression is $x= 0.046m$

Explanation:

From the question we are told that

The mass of the block is $m_b = 1.45 kg$

The spring constant is $k = 860 N/m$

The coefficient of static friction is $\mu = 0.36$

For the the block not slip it mean the sum of forces acting on the horizontal axis is equal to the forces acting on the vertical axis

Now the force acting on the vertical axis is the force due to gravity which is mathematically given as

$F_y = m_b*g$

And the force acting on the horizontal axis is force due to the spring which is mathematically represented as

$F_x = k *x * \mu$

where x is the minimum compression to keep the block from slipping

Now equating this two formulas and making x the subject

$x = \frac{m_b * g}{k * \mu}$

substituting values we have

$x = \frac{1.45 * 9.8}{860 *0.36}$

$x= 0.046m$

3 0

3 years ago

From what does oil form?

Akimi4 [234]

C dinosaurs. Once they break down and get pressurized they turn to oil

6 0

3 years ago