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3 years ago

A train traveling at 6.4 m/s accelerates at 0.10 m/s 2 over a distance of 100 m. How large is the train’s final velocity?

Physics

1 answer:

klasskru [66]3 years ago

4 0

The final velocity of the train at the end of the given distance is 7.81 m/s.

The given parameters;

initial velocity of the train, u = 6.4 m/s
acceleration of the train, a = 0.1 m/s²
distance traveled, s = 100 m

The final velocity of the train at the end of the given distance is calculated using the following kinematic equation;

v² = u² + 2as

v² = (6.4)² + (2 x 0.1 x 100)

v² = 60.96

v = √60.96

v = 7.81 m/s

Thus, the final velocity of the train at the end of the given distance is 7.81 m/s.

Learn more here:brainly.com/question/21180604

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A force of 20. Newtons to the left exerted on a cart for 10. Seconds. For what period of time must a 50.-newton force to the rig

FinnZ [79.3K]

Impulse = (force) x (time)

The first impulse was (20 N) x (10 sec) = 200 meters/sec

The second one is (50 N) x (time) and we want it equal to the first one, so

(50 N) x (time) = 200 meters/sec

Divide each side by 50N : Time = 200/50 = <em>4 seconds</em>

By the way, the quantity we're playing with here is the cart's <em>momentum</em>.

6 0

3 years ago

f the absolute pressure in a tank is 128 kPa , determine the pressure head in mm of mercury. The atmospheric pressure is 100 kPa

White raven [17]

Answer:

211 mmHg

Explanation:

Absolute Pressure = Gauge Pressure + Atmospheric pressure

128 = Gauge Pressure + 100

Gauge Pressure = 28 KPa = 28 × 10³ Pa

Also Gauge Pressure = ρgh

ρ = density = 13550 kg/m³

g = acceleration due to gravity = 9.8 m/s²

h = pressure head = ?

28 × 10³ = 13550 × 9.8 × h

h = 28000/(13550 × 9.8)

h = 0.211 m = 211 mm

5 0

3 years ago

A bus driver drove from Philadelphia to Washington DC. He drove the first 100km in 2 hours, the next 55km in 1 hour, and the fin

Alla [95]

Answer:

b. 46 km/hr

Explanation:

Applying,

S' = d'/t'................... Equation 1

Where S' = Average speed of the bus, d' = Total distance covered by the bus, t' = Total time taken.

From the question,

d' = (100+55+75) km = 230 km

t' = (2+1+2) = 5 hours

Substitute these values into equation 1

S' = 230/5

S' = 46 km/hr.

Hence the correct option is b. 46 km/hr

8 0

3 years ago

If an astronaut can jump straight up to a height of 0.6 m on earth, how high could he jump on the moon?

Zinaida [17]

Answer:

Explanation:

Given

Person on earth can jump to a height $s=0.6\ m$

initial velocity is u

using

$v^2-u^2=2as$

where v=final velocity

u=initial velocity

a=acceleration

s=displacement

final velocity is zero

$s=\frac{u^2}{2g}$

$0.6=\frac{u^2}{2g}-----1$

On moon surface acceleration due to gravity is \frac{1}{6}[/tex] th of earth gravity

so height attained is given by

$h=\frac{u^2\times 6}{2\cdot g}-----2$

divide 1 and 2 we get

$h=6\times 0.6=3.6\ m$

4 0

3 years ago

The initial temperature of 150 g of ice is ????20°C. The spe- cific heat capacity of ice is 0.5 cal/g·C° and water’s is 1 cal/g·

soldier1979 [14.2K]

1. 13,500 cal

First of all, we need to find the amount of heat needed to raise the temperature of the ice from -20°C to 0°C. This is given by

$Q_1 = m C_i \Delta T$

where

m = 150 g is the mass of the ice

C_i = 0.5 cal/g·C° is the specific heat capacity of the ice

$\Delta T=0 C-(-20 C)=20^{\circ}C$ is the change in temperature of the ice

Substituting,

$Q_1 = (150 g)(0.5 cal/gC)(20 C)=1500 cal$

Now we have to find the amount of heat needed to melt the ice, which is

$Q_2 = m \lambda_f$

where

m = 150 g is the mass of the ice

$\lambda_f = 80 cal/g$ is the latent heat of fusion

Substituting,

$Q_2 = (150 g)(80 cal/g)=12,000 cal$

So the total heat required is

$Q_3 = 1500 cal + 12,000 cal = 13,500 cal$

2. 3750 cal

The additional amount of heat required to heat the water to 25°C is

$Q_4 = m C_w \Delta T$

where

m = 150 g is the mass of water

C_w = 1 cal/g·C is the speficic heat capacity of water

$\Delta T=25 C-0 C=25^{\circ}C$ is the change in temperature

Substituting,

$Q_4 = (150 g)(1 cal/gC)(25 C)=3,750 cal$

3. 9200 cal

First of all, we need to find the amount of heat needed to raise the temperature of the ice from -20°C to 0°C. As at point 1., this is given by

$Q_1 = m C_i \Delta T$

where

m = 80 g is the mass of the ice

C_i = 0.5 cal/g·C° is the specific heat capacity of the ice

$\Delta T=0 C-(-20 C)=20^{\circ}C$ is the change in temperature of the ice

Substituting,

$Q_1 = (80 g)(0.5 cal/gC)(20 C)=800 cal$

Now we have to find the amount of heat needed to melt the ice:

$Q_2 = m \lambda_f$

where

m = 80 g is the mass of the ice

$\lambda_f = 80 cal/g$ is the latent heat of fusion

Substituting,

$Q_2 = (80 g)(80 cal/g)=6,400 cal$

Finally, the amount of heat required to heat the water to 25°C is

$Q_3 = m C_w \Delta T$

where

m = 80 g is the mass of water

C_w = 1 cal/g·C is the speficic heat capacity of water

$\Delta T=25 C-0 C=25^{\circ}C$ is the change in temperature

Substituting,

$Q_3 = (80 g)(1 cal/gC)(25 C)=2,000 cal$

So the total heat required is

$Q=Q_1+Q_2+Q_3=800 cal+6,400 cal+2,000 cal=9,200 cal$

4. No

Explanation:

The total heat required for this process consists of 3 different amounts of heat:

1- The heat required to bring the ice at melting temperature

2- The heat required to melt the ice, while its temperature stays constant

3- The heat required to raise the temperature of the water

However, computing how much heat is required to melt the ice and adding the amount of heat required to raise the temperature of 80 g of water by 45°C is not equivalent: in fact, the calculation of point 1) requires to use the specific heat capacity of ice, not that of water, therefore the two are not equivalent.

4 0

3 years ago