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3 years ago

A train starts from rest with acceleration of 0.5ms-2 find speed in Kmh-1 when it moved through 100m. (36kmh-1)

Physics

1 answer:

SashulF [63]3 years ago

4 0

Answer:

36km/h

Explanation:

since the acceleration of the train was 0.5m/s² i.e it is constant . since the acceleration is constant you can use one of the "SUVAT" equations to find the answer.

the equation that best fits this question is: v²=u² + 2āS.

v stands for final velocity

u stands for intial velocity

v²=u² + 2āS since the car started from rest the intial velocity is zero.

v²=u² + 2āS

v²= 0 + 2(0.5m/s²)*100m

v²= 1m/s²*100m

√v²=√100m²/s²

v=10m/s

1km/hr = 1/3.6m/s

x. = 10m/s.

use criss cross method to find the value of x

x=( 1km/hr* 10m/s)

1/3.6m/s

x= 36km/hr

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Two metallic rods A and B of different materials have same length. The linear expansivity of A is 12×10–6 oC–1and cubical expans

uysha [10]

Answer:

The length of rod A will be greater than the length of rod B

Explanation:

We, know that the formula for final length in linear thermal expansion of a rod is:

L' = L(1 + ∝ΔT)

where,

L' = Final Length

L = Initial Length

∝ = Co-efficient of linear expansion

ΔT = Change in temperature

Since, the rods here have same original length and the temperature difference is same as well. Therefore, the final length will only depend upon the coefficient of linear expansion.

For Rod A:

∝₁ = 12 x 10⁻⁶ °C⁻¹

For Rod B:

∝₂ = β₂/3

where,

β₂ = Coefficient of volumetric expansion for rod B = 24 x 10⁻⁶ °C⁻¹

Therefore,

∝₂ = 24 x 10⁻⁶ °C⁻¹/3

∝₂ = 8 x 10⁻⁶ °C⁻¹

Since,

∝₁ > ∝₂

Therefore,

L₁ > L₂

So, the length of rod A will be greater than the length of rod B

6 0

3 years ago

A projectile is launched diagonally into the air and has a hang time of 24.5 seconds. Approximately how much time is required fo

Rasek [7]

Answer:

$t=12.25\ seconds$

Explanation:

Diagonal Launch

It's referred to as a situation where an object is thrown in free air forming an angle with the horizontal. The object then describes a known path called a parabola, where there are x and y components of the speed, displacement, and acceleration.

The object will eventually reach its maximum height (apex) and then it will return to the height from which it was launched. The equation for the height at any time t is

$x=v_ocos\theta t$

$\displaystyle y=y_o+v_osin\theta \ t-\frac{gt^2}{2}$

Where vo is the magnitude of the initial velocity, $\theta$ is the angle, t is the time and g is the acceleration of gravity

The maximum height the object can reach can be computed as

$\displaystyle t=\frac{v_osin\theta}{g}$

There are two times where the value of y is $y_o$ when t=0 (at launching time) and when it goes back to the same level. We need to find that time t by making $y=y_o$

$\displaystyle y_o=y_o+v_osin\theta\ t-\frac{gt^2}{2}$

Removing $y_o$ and dividing by t (t different of zero)

$\displaystyle 0=v_osin\theta-\frac{gt}{2}$

Then we find the total flight as

$\displaystyle t=\frac{2v_osin\theta}{g}$

We can easily note the total time (hang time) is twice the maximum (apex) time, so the required time is

$\boxed{t=24.5/2=12.25\ seconds}$

4 0

3 years ago

A student lifts a 90 pound (lb) ball 4 feet (ft) in 5 seconds (s). How many joules of work has the student completed?

scZoUnD [109]

Wending your way through this question and the list of choices
is like running through a corn-maze on a moonless night with
your eyes closed.

Work = (force) x (distance)

Keeping with the units used In the question ...

   Work = (90 pounds) x (4 feet) = 360 foot-pounds .

If you want the work expressed in joules, then you need to convert
pounds and feet to newtons and meters.

   (90 pounds) x (4.448 newtons/pound) = 400.32 Newtons

   (4 feet) x (1 meter / 3.28084 feet) = 1.219 meters

Work = (force) x (distance)

      = (400.32 newtons) x (1.219 meters) = 488.1 joules .

7 0

3 years ago

A 110-kg object and a 410-kg object are separated by 3.80 m.

aniked [119]

Answer:

a) Fₙ = 2,273 10⁻⁷ N and b) x₃ = 1,297 m

Explanation:

This problem can be solved using the law of universal gravitation and Newton's second law for the equilibrium case. The Universal Gravitation Equation is

F = G m₁ m₂ / r₁₂²

a) we write Newton's second law

Σ F = F₁₃ - F₃₂

Body 1 has mass of m₁ = 110 kg and we will place our reference system, body 2 has a mass of m₂ = 410 kg and is in the position x₂ = 3.80 m

Body 3 has a mass of m₃ = 41.0 kg and is in the middle of the other two bodies

x₃ = (x₂-x₁) / 2

x₃ = 3.80 / 2 = 1.9 m

Fₙ = -G m₁ m₃ / x₃² + G m₃ m₂ / x₃²

Fₙ = G m₃ / x₃² (-m₁ + m₂)

Calculate

Fₙ = 6.67 10⁻¹¹ 41.0 / 1.9² (- 110 + 410)

Fₙ = 2,273 10⁻⁷ N

Directed to the right

b) find the point where the force is zero

The distance is

x₁₃ = x₃ - 0

x₃₂ = x₂ -x₃= 3.8 -x₃

We write the park equation net force be zero

0 = - F₁₃ + F₃₂

F₁₃ = F₃₂

G m₁ m₃ / x₁₃² = G m₃ m₂ / x₃₂²

m₁ / x₁₃² = m₂ / x₃₂²

Let's look for the relationship between distances, substituting

m₁ / x₃² = m₂ / (3.8 - x₃)²

(3.8 - x₃) = x₃ √ (m₂ / m₁)

x₃ + x₃ √ (m₂ / m₁) = 3.8

x₃ (1 + √ m₂ / m₁) = 3.8

x₃ = 3.8 / (1 + √ (m₂ / m₁))

x₃ = 3.8 / (1 + √ (410/110))

x₃ = 1,297 m

When body 3 is in this position the net force on it is zero

8 0

3 years ago

I need this ASAP will give brainliest Question 2

myrzilka [38]

Answer:

2a. 24 m³

2b. 19200 Kg

2c. 192000 N

2d. 16000 N/m²

Explanation:

2a. Determination of the volume of paraffin.

Length (L) = 4 m

Width (W) = 3 m

Height (H) = 2 m

Volume (V) =?

V = L × W × H

V = 4 × 3 × 2

V = 24 m³

Therefore the volume of paraffin is 24 m³

2b. Determination of the mass of paraffin.

Volume (V) = 24 m³

Density (ρ) = 800 Kg/m³

Mass (m) =?

ρ = m / V

800 = m / 24

Cross multiply

m = 800 × 24

m = 19200 Kg

Therefore, the mass of paraffin is 19200 Kg

2c. Determination of the weight of paraffin.

Mass (m) = 19200 Kg

Acceleration due to gravity (g) = 10 N/Kg

Weight (W) =?

W = m × g

W = 19200 × 10

W = 192000 N

Therefore, the weight of paraffin is 192000 N

2d. Determination of the pressure.

Density (ρ) = 800 Kg/m³

Depth (h) = 2 m

Acceleration due to gravity (g) = 10 N/Kg

Pressure (P) =?

P = ρgh

P = 800 × 10 × 2

P = 16000 N/m²

4 0

3 years ago