What is a molecule called when it has more than one element?

Which of the following positions would Earth be in if it were winter in the Northern Hemisphere?

abruzzese [7]

Answer:

Earth's tilted axis causes the seasons. Throughout the year, different parts of Earth receive the Sun's most direct rays. So, when the North Pole tilts toward the Sun, it's summer in the Northern Hemisphere. And when the South Pole tilts toward the Sun, it's winter in the Northern Hemisphere.

6 0

3 years ago

A block of mass 10 kg and measuring 250 mm on each edge is pulled up an inclined surface on which there is a film of SAE 10W-30

Ivan

Answer:

a) 2.53 * 10^-2 m/s

b) -4.78 * 10^-2 m/s

c) 1.21 * 10^-1 m/s

Explanation:

Given data :

Mass of block = 10 kg

Measuring 250mm on each side

a) calculate the speed when a force of 75N is applied to pull block upwards

F = f + W sin∅ ( equation for applying the force of equilibrium condition in the x axis ) ----- ( 1 )

f ( friction force )= ( 16400v * 6.25 *10^-2) = 1025 v

F ( force applied ) = 75

W ( weight of block ) = 10 * 9.81 = 98.1 N

∅ = 30°

input values into equation 1

V = $\frac{75- (98.1*sin30^{0}) }{1025}$ = 2.53 * 10^-2 m/s

b) Speed when no force is applied on the block

F = f + W sin∅

F = 0

f = 1025 V

W = 98.1 N

∅ = 30°

hence V = $\frac{0 - (98.1*sin30^{0}) }{1025}$ = - 4.78 * 10^-2 m/s

c) when a force is applied to push block down the incline

F = f + W sin∅ ----- ( 3 )

F = 75 N

f = 1025 V

W = 98.1 N

∅ = 30°

input values into equation 3 considering the fact that the weight of the block is acting in the opposite direction

75 = 1025 V - 98.1 ( sin 30° )

V = $\frac{75+( 98.1*sin30^{0}) }{1025}$ = 1.21 * 10^-1 m/s

5 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:

<em>a. The rock takes 2.02 seconds to hit the ground</em>

<em>b. The rock lands at 20,2 m from the base of the cliff</em>

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

3 years ago

A stone thrown horizontally from a height of 5.72 m hits the ground at a distance of 13.30 m. Calculate the initial speed of the

kakasveta [241]

Answer:

initial velocity=12.31 m/s

Final speed= 16.234 m/s

Explanation:

Given Data

height=5.72 m

distance=13.30 m

To Find

Initial Speed=?

Solution

Use the following equation to determine the time of the stone is falling.

d = vi ×t ½ ×9.8 × t²

Where

d = 5.72m and vi = 0 m/s

so

5.72 = ½× 9.8 ×t²

t = √(5.72 ÷ 4.9)

t=1.08 seconds

To determine the initial horizontal velocity use the following equation.

d = v×t

13.30 = v ×1.08

v = 13.30 ÷ 1.08

v=12.31 m/s

To determine stone’s final vertical velocity use the following equation

vf = vi+9.8×t............vi=0 m/s

vf = 9.8×1.08

vf= 10.584 m/s

To determine stone’s final speed use the following equation

Final speed = √[Horizontal velocity²+Final vertical velocity²]

Final speed = √{(12.31 m/s)²+(10.584 m/s)²}

Final speed= 16.234 m/s

3 0

4 years ago

What is the upper block's acceleration if the coefficient of kinetic friction between the block and the table is 0.13?

IgorLugansk [536]

Let us assume that pulley is mass less.

Let the tension produced at both ends of the pulley is T.

We are asked to calculate the acceleration of the block.

Let the masses of two bodies are denoted as $m_{1} \ and\ m_{2}\ respectively$

$Let\ m_{1} =1 kg\ and\ m_{2} =2 kg$

As per this diagram, the body having mass 1 kg is moving downward and the body having mass 2 kg is moving on the surface of the table.

Let the acceleration of each block is a .

For body having mass 1 kg:

The net force acting on 1 kg body will be-

$m_{1} g-T=m_{1} a$ [1]

Here tension in the rope will be vertically upward and weight of the body will be in vertical downward direction.

For body having mass 2 kg:

The coefficient of kinetic friction $[\mu]=0.13$

$Hence\ the\ frictional\ force\ F=\mu N$

$F=\mu m_{2} g$

Hence the net force acting on the body having mass 2 kg-

$T-\mu m_{2} g=m_{2} a$ [2]

Here the tension of the rope is towards right i.e along the direction of motion of the 2 kg block and frictional force is towards left.

Combining 1 and 2 we get-

$m_{1} g-T=m_{1}a$ [1]

$T-\mu m_{2}g= m_{2} a$ [2]

---------------------------------------------------

$[m_{1} -\mu m_{2} ]g=[m_{1} +m_{2} ]a$

$a=\frac{m_{1}-\mu m_{2}} {m_{1}+ m_{2}}*g$

$a=\frac{1-[2*0.13]}{1+2} *9.8\ m/s^2$

$a=\frac{0.74}{3} *9.8\ m/s^2$

$a=2.417 m/s$ [ans]

6 0

3 years ago

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