How do you use the SI conversion tool to perfom metric conversions?

if you crash your car how could you decrease the damage to you or the car using the concept of impulse

kotykmax [81]

Explanation:

Crumple zones are sections in cars that are designed to crumple up when the car encounters a collision. Crumple zones minimize the effect of the force in an automobile collision in two ways. By crumpling, the car is less likely to rebound upon impact, thus minimizing the momentum change and the impulse.

3 0

3 years ago

A garden hose having with an internal diameter of 1.1 cm is connected to a (stationary) lawn sprinkler that consists merely of a

trapecia [35]

Answer:

Water leaves the sprinkler at a speed of 2.322 m/sec

Explanation:

We have given internal diameter of the garden $d_1=1.1cm$

Speed of water in the hose is $v_1=0.95m/sec$

Number of holes n = 22

Diameter of each holes $d_2=15cm$

According to continuity equation $A_1v_1=A_2v_2$

$d_1^2\times v_1=22\times d_2^2v_2$

$1.1^2\times 0.95=22\times 0.15^2\times v_2$

$v_2=2.322m/sec$

So water leaves the sprinkler at a speed of 2.322 m/sec

6 0

3 years ago

an object traveling 200 feet per second slows to 50 feet per second in 5 seconds. Calculate the acceleration of the object

emmainna [20.7K]

Answer:

The acceleration of the object is $-30\ m/s^2$

Explanation:

Given:

Initial velocity of object $v_i$ = 200 feet/second

Final velocity of object $v_f$ = 50 feet/second

Time of travel = 5 seconds

To calculate acceleration of the object we will find the rate of change of velocity with respect to time.

So, acceleration $a$ is given by:

$a=\frac{v_f-v_i}{t}$

where $v_f$ represents final velocity, $v_i$ represents initial velocity and $t$ is time of travel.

Plugging in values to evaluate acceleration.

$a=\frac{50-200}{5}$

$a=\frac{-150}{5}$

$a=-30\ m/s^2$

The acceleration of the object is $-30\ m/s^2$ (Answer). The negative sign shows the object is slowing down.

4 0

4 years ago

The Shinkansen (bullet train in Japan) makes a trip from Tokyo Station to Kyoto station in 2 hours and 14 min. The distance trav

11Alexandr11 [23.1K]

Answer:

v = 57.2 m/s

Explanation:

The average velocity of the train can be defined as the total distance covered by the train divided by the time taken by the train to cover that distance. Therefore, we will use the following formula to find the average velocity of the train:

v = s/t

where,

s = distance covered = 460 km = (460 km)(1000 m/1 km) = 4.6 x 10⁵ m

t = time taken to cover the distance = 2 h 14 min

Now, we convert it into minutes:

t = (2 h)(60 min/1 h) + 14 min

t = 120 min + 14 min = (134 min)(60 s/1 min)

t = 8040 s

Therefore, the value of velocity will be:

v = (4.6 x 10⁵ m)/8040 s

7 0

3 years ago

In which of the two situations described is more energy transferred?

Furkat [3]

Answer:

More energy is transferred in situation A

Explanation:

Each of the situations are analyzed as follows;

Situation A

The temperature of the cup of hot chocolate = 40 °C

The temperature of the interior of the freezer in which the chocolate is placed = -20 °C

We note that at 0°C, the water in the chocolate freezes

The energy transferred by the chocolate to the freezer before freezing is given approximately as follows;

E₁ = m×c₁×ΔT₁

Where;

m = The mass of the chocolate

c₁ = The specific heat capacity of water = 4.184 kJ/(kg·K)

ΔT₁ = The change in temperature from 40 °C to 0°C

Therefore, we have;

E₁ = m×4.184×(40 - 0) = 167.360·m kJ

The heat the coffee gives to turn to ice is given as follows;

E₂ = m· $H_f$

Where;

$H_f$ = The latent heat of fusion = 334 kJ/kg

∴ E₂ = m × 334 kJ/kg = 334·m kJ

The heat required to cool the frozen ice to -20 °C is given as follows;

E₃ = m·c₂·ΔT₂

Where;

c₂ = The specific heat capacity of ice = 2.108 kJ/(kg·K)

Therefore, we have;

E₃ = m × 2.108 ×(0 - (-20)) = 42.16

E₃ = 42.16·m kJ/(kg·K)

The total heat transferred = (167.360 + 334 + 42.16)·m kJ/(kg·K) = 543.52·m kJ/(kg·K)

Situation B

The temperature of the cup of hot chocolate = 90 °C

The temperature of the room in which the chocolate is placed = 25 °C

The heat transferred by the hot cup of coffee, E, is given as follows;

E = m×4.184×(90 - 25) = 271.96

∴ E = 271.96 kJ/(kg·K)

Therefore, the total heat transferred in situation A is approximately twice the heat transferred in situation B and is therefore more than the heat transferred in situation B

Energy transferred in situation A = 543.52 kJ/(kg·K)

Energy transferred in situation B = 271.96 kJ/(kg·K)

Energy transferred in situation A ≈ 2 × Energy transferred in situation B

∴ Energy transferred in situation A > Energy transferred in situation B.

3 0

3 years ago