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3 years ago

If a train travels at a speed of 40 km/h for 3 hours, how far will it travel

Physics

2 answers:

VikaD [51]3 years ago

8 0

120 km/3 hours. 40/1=?/3 1x3=3 hours so 40x3=120 km

Alik [6]3 years ago

4 0

Answer:

120 meters

Explanation:

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velikii [3]

7) 6, i believe, (Cu) 1 atom+ (S) 1 atom+(0)4 atoms.

7 0

3 years ago

A 51.0 kg crate, starting from rest, is pulled across a floor with a constant horizontal force of 225 N. For the first 10.0 m th

shepuryov [24]

Answer:

The final speed of the crate is 12.07 m/s.

Explanation:

For the first 10.0 meters, the only force acting on the crate is 225 N, so we can calculate the acceleration as follows:

$F = ma$

$a = \frac{F}{m} = \frac{225 N}{51.0 kg} = 4.41 m/s^{2}$

Now, we can calculate the final speed of the crate at the end of 10.0 m:

$v_{f}^{2} = v_{0}^{2} + 2ad_{1}$

$v_{f} = \sqrt{0 + 2*4.41 m/s^{2}*10.0 m} = 9.39 m/s$

For the next 10.5 meters we have frictional force:

$F - F_{\mu} = ma$

$F - \mu mg = ma$

So, the acceleration is:

$a = \frac{F - \mu mg}{m} = \frac{225 N - 0.17*51.0 kg*9.81 m/s^{2}}{51.0 kg} = 2.74 m/s^{2}$

The final speed of the crate at the end of 10.0 m will be the initial speed of the following 10.5 meters, so:

$v_{f}^{2} = v_{0}^{2} + 2ad_{2}$

$v_{f} = \sqrt{(9.39 m/s)^{2} + 2*2.74 m/s^{2}*10.5 m} = 12.07 m/s$

Therefore, the final speed of the crate after being pulled these 20.5 meters is 12.07 m/s.

I hope it helps you!

7 0

3 years ago

A loaf of bread (volume 3100 cm3) with a density of 0.90 g/cm3 is crushed in the bottom of the grocery bag into a volume of 1240

andrew11 [14]

Answer:

2,25 g/cm3

Explanation:

Hi, you have to know one thing for this.. Density = mass/Volume,

When you have the loaf of bread with 3100 cm3 and a density of 0.90 g/cm3, the mass of that bread is 2790 g because of if you isolate the variable mass from the equation you get.. mass= density x volume

Later, have on account the mass never changes, so you crush the bread and the mass is the same.. so when you have the mashed bread.. you know that the mass is 2790 g and the volume of the bag is 1240 cm3, so you apply the main equation.... density=2790 g / 1240 cm3 , so density = 2,25 g/cm3

8 0

3 years ago

the distance a spring will stretch varies directly with how much weight is attached to the spring. If a spring stretches 11 inch

Paha777 [63]

Direct variation involves ration and proportions, so

you need to set up the proportion:

<span>11 / 75 = x / 65

Cross multiplying:

75x = 11*65

x = (11*65)/75

Solving, we get x = 9.533, </span>

<span>which rounds off to 9.5

Therefore, the spring will stretch up to 9.5 inches with 65 attached.

I hope my answer has come to your help. Thank you for posting your question here in Brainly.

</span>

4 0

3 years ago

consider a sample of gas in a container on a comfortable spring day. the Celsius temperature suddenly doubles, and you transfer

Tasya [4]

Answer:

6.6 atm

Explanation:

Using the general gas law

P₁V₁/T₁ = P₂V₂/T₂

Let P₂ be the new pressure

So, P₂ = P₁V₁T₂/V₂T₁

Since V₂ = 2V₁ , P₁ = 12 atm and T₁ = 273 + t where t = temperature in Celsius

T₂ = 273 + 2t (since its Celsius temperature doubles).

Substituting these values into the equation for P₂, we have

P₂ = P₁V₁(273 + 2t)/2V₁(273 + t)

P₂ = 12(273 + 2t)/[2(273 + t)]

P₂ = 6(273 + 2t)/(273 + t)]

assume t = 30 °C on a comfortable spring day

P₂ = 6(273 + 2(30))/(273 + 30)]

P₂ = 6(273 + 60))/(273 + 30)]

P₂ = 6(333))/(303)]

P₂ = 6.6 atm

7 0

3 years ago