If a bird travels 40 km/h to 80 km/h in 2 s in the air, what is the acceleration 5 points

A ball rolls at a speed of 6cm/s. How far does the ball roll in 40 seconds ? ____cm

sesenic [268]

Answer:

240cm

Explanation:

Speed = 6cm/s

Time = 40 seconds

Speed= distance/time

6cm/s = distance/40seconds

Distance = 6×40

Distance = 240cm

Hence, in 40 seconds, the ball must have rolled the distance of 240cm

8 0

3 years ago

A fireboat is to fight fires at coastal areas by drawing seawater with a density of 1030 kg/m3 through a 10-cm-diameter pipe at

GaryK [48]

Answer:

50.93 m/s

199.5 kW

Explanation:

From the question, the nozzle exit diameter = 5 cm, Radius= diameter/2= 5cm/2= 2.5cm. we can convert it to metre for unit consistency= (2.5×0.01)=

0.025m

We can calculate the The cross sectional area of the nozzle as

A= πr^2

A= π ×0.025^2

= 1.9635 ×10^- ³ m²

From the question, the water is moving through the pipe at a rate of 0.1 m /s , then for the water to move through it at a seconds, it must move at

(0.1 / 1.9635 ×10^- ³ m²)

= 50.93 m/s

During the Operation of the pump, the Dynamic energy of the water= potential energy provided there is no loss during the Operation

mgh = 1/2mv²

We can make "h" subject of the formula, which is the height of required head of water

h = (1/2mv²)/mg

h= v² / 2g

h = 50.93² / (2 ×9.81)

h = 132.21m

From the question;

The total irreversible head loss of the system = 3 m,

the given position of nozzle = 3 m

the total head the pump needed=(The total irreversible head loss of the system + the position of the nozzle + required head of water )

=(3 + 3 + 132.21m)

=138.21m

mass of water pumped in a seconds can be calculated since we know that mass is a product of volume and density

Volume= 0.1m³

Density of sea water=1030 kg/m

(0.1 m^3× 1030)

= 103kg

We can calculate the Potential enegry, which is = mgh

= (103 ×9.81 × 138.21)

= 139651.5 Watts

= 139.65kW

To determine required shaft power input to the pump and the water discharge velocity

Energy= efficiency × power

But we are given efficiency of 70 percent, then

139651.5 Watts = 0.7P

=199502.18 Watts

P=199.5 kW

Therefore, the required shaft power input to the pump and the water discharge velocity is 199.5 kW

5 0

2 years ago

The heat capacity of nickel is 0.444 J/(g · °C). Calculate the amount of heat needed to raise the temperature of 18 g of nickel

azamat

The final speed of the nickel at the given quantity of heat is determined as 202.1 m/s.

<h3>Final speed of the nickel</h3>

Apply the principle of conservation of energy.

Q = mcΔθ

Q = (18)(0.444)(66 - 20)

Q = 367.63 J

Q = K.E = ¹/₂mv²

2K.E = mv²

v = √(2K.E/m)

where;

v is the final speed

v = √(2 x 367.63)/(0.018))

v = 202.1 m/s

Learn more about speed here: brainly.com/question/4931057

#SPJ1

5 0

1 year ago

An object of mass 30 kg is falling in air and experiences a force due to air resistance of 50 newtons. Calcuate the acceleration

rjkz [21]

Answer:

this answer you question

6 0

3 years ago

Two blocks of masses M 1 and M 2 are connected by a massless string that passes over a massless pulley as shown in the figure. M

stealth61 [152]

The mass M1 is 7.8 kg

Explanation:

Block M1 is hanging on the string while block M2 is on the frictionless ramp.

We have to write the equations of motion for the two blocks.

- For M1, the only two forces acting on it are the force of gravity $M_1 g$ (downward) and the tension in the string T (upward). So we can write

$M_1 g - T = M_1 a$

where

$M_1$ is the mass of the block

$g=9.8 m/s^2$ is the acceleration of gravity

$a$ is the acceleration of the system

- For M2, the only two forces acting on it are the tension in the string T (acting up along the ramp) and the component of the gravity acting down along the ramp, $M_2 g sin \theta$ . So the equation of motion is

$T-M_2 g sin \theta = M_2 a$

where

$M_2 = 13.5 kg$ is the mass of the 2nd block

$\theta=35.5^{\circ}$ is the angle of the ramp

In order for the two blocks to be in equilibrium, the acceleration must be zero:

$a=0$

So the two equations become:

$M_1 g - T=0\\T-M_2 g sin \theta = 0$

Isolating T from the 1st equation,

$T=M_1 g$

And substituting into the 2nd equation, we can find the value of the mass $M_1$ :

$M_1 g - M_2 g sin \theta = 0\\M_1 = M_2 sin \theta = (13.5)(sin 35.5^{\circ})=7.8 kg$

Learn more about acceleration and forces:

brainly.com/question/11411375

brainly.com/question/1971321

brainly.com/question/2286502

brainly.com/question/2562700

#LearnwithBrainly

7 0

3 years ago