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marishachu [46]

2 years ago

10

a high speed train travels with an average speed of 250 km/h. the train travels for 2 hrs. how far does the train travel

2 answers:

Afina-wow [57]2 years ago

7 0

Answer:

The answer would be 454 miles.

Explanation:

You multiply the average speed by time travelled. Hope this helps!

steposvetlana [31]2 years ago

5 0

Answer:

<h3>The answer is 500 km </h3>

Explanation:

The distance covered by an object given it's velocity and time taken can be found by using the formula

<h3>distance = average velocity × time</h3>

From the question

average speed = 250 km/h

time = 2 hrs

We have

distance = 250 × 2

We have the final answer as

<h3>500 km</h3>

Hope this helps you

You might be interested in

If you were to drop a rock from a tall building, assuming that it had not yet hit the ground, and neglecting air resistance, aft

elena-14-01-66 [18.8K]

Since the object is dropped from some height so its initial speed must be zero

acceleration of the object is due to gravity

so we can use kinematics to find the time it will take to drop by x = 22 m

$\delta x = v_i * t + \frac{1}{2}at^2$

$22 = 0 + \frac{1}{2}*9.8*t^2$

$t = 2.12 s$

Now the speed after 2.12 s will be given as

$v_f = v_i + at$

$v_f = 0 + 9.8 * 2.12$

$v_f = 20.8 m/s$

so above is the speed and time

5 0

2 years ago

A child is trying to throw a ball over a fence. She gives the ball an initial speed of 8.0 m/s at an angle of 40° above the hori

EastWind [94]

Answer:

the child is 1.581 m far from the fence

Explanation:

The diagrammatic illustration that give a better view of what the question denote can be seen in the image attached below.

From the image attached below, let assume that the release point is the origin, then equation of the motion (x) is as follows:

$x - x_o = u_xt$

$\mathtt{x = u_xt \ \ \ since (x_o = 0)}$ ---- (1)

the equation of the motion y is :

$\mathtt{y - y_o =u_yt - 0.5 gt^2}$

$\mathtt{y = u_yt-4.9t^2 \ \ \ since (y_o =0)}$

$\mathtt{ 1= (u \ sin 40^0)t -4.9 \ t^2 }$

$\mathtt{1 = 8 sin 40^0 t - 4.9 t^2}$

$\mathtt{1 = 5.14t - 4.9t^2}$

$\mathtt{4.9t^2 - 5.14t +1 = 0}$

By using the quadratic formula, we have;

$\mathtt{ \dfrac{ -b \pm \sqrt{b^2 - 4ac}}{2a}} }$

where;

a = 4.9, b = -5.14 c = 1

$= \mathtt{ \dfrac{ -(-5.14) \pm \sqrt{(-5.14)^2 - 4(4.9)(1)}}{2(4.9)}} }$

$= \mathtt{ \dfrac{ 5.14 \pm \sqrt{26.4196 -19.6}}{9.8}} }$

$= \mathtt{ \dfrac{ 5.14 \pm \sqrt{6.8196}}{9.8}} }$

$= \mathtt{ \dfrac{ 5.14+ \sqrt{6.8196}}{9.8} \ \ OR \ \ \dfrac{ 5.14- \sqrt{6.8196}}{9.8}} }$

$= \mathtt{ \dfrac{ 5.14+ 2.6114}{9.8} \ \ OR \ \ \dfrac{ 5.14- 2.6114}{9.8}} }$

$= \mathtt{ \dfrac{ 7.7514}{9.8} \ \ OR \ \ \dfrac{ 2.5286}{9.8}} }$

$= \mathbf{ 0.791 \ \ OR \ \ 0.258} }$

In as much as the ball is traveling upward, then we consider t= 0.258sec.

From equation (1)

$\mathtt{x = u_x(0.258)}$

$\mathtt{x = ucos 40^0 (0.258)}$

$\mathtt{x = 8 \ cos 40^0 (0.258)}$

$\mathbf{x = 1.581 \ m}$

Thus, the child is 1.581 m far from the fence

6 0

3 years ago

(Figure 1) shows the angular-velocity-versus-time graph for a particle moving in a circle. How many revolutions does the object

Travka [436]

Answer: 10.34

Explanation:

Given

$\omega -t$ graph for a particle is given

angle turned by the particle in radians is given by the area under $\omega -t$ graph

The area is given by

$A=20\times (2-0)+10(4-2)+\dfrac{1}{2}\times (20-10)\times (3-2)\\A=40+20+5=65\ rad$

Revolutions(N) made by the object is given by

$N=\dfrac{65}{2\pi }=10.34$

4 0

2 years ago

A light bulb does 100 joules of work in 2.5 seconds. how much work power dos it have

Lesechka [4]

100/2.5 because power=energy/time

5 0

3 years ago

A 50-g cube of ice, initially at 0.0°C, is dropped into 200 g of water in an 80-g aluminum container, both initially at 30°C.

MakcuM [25]

Answer:

b. 9.5°C

Explanation:

$m_i$ = Mass of ice = 50 g

$T_i$ = Initial temperature of water and Aluminum = 30°C

$L_f$ = Latent heat of fusion = $3.33\times 10^5\ J/kg^{\circ}C$

$m_w$ = Mass of water = 200 g

$c_w$ = Specific heat of water = 4186 J/kg⋅°C

$m_{Al}$ = Mass of Aluminum = 80 g

$c_{Al}$ = Specific heat of Aluminum = 900 J/kg⋅°C

The equation of the system's heat exchange is given by

$m_i(L_f+c_wT)+m_wc_w(T-T_i)+m_{Al}c_{Al}=0\\\Rightarrow 0.05\times (3.33\times 10^5+4186\times T)+0.2\times 4186(T-30)+0.08\times 900(T-30)=0\\\Rightarrow 1118.5T-10626=0\\\Rightarrow T=\dfrac{10626}{1118.5}\\\Rightarrow T=9.50022\ ^{\circ}C$

The final equilibrium temperature is 9.50022°C

4 0

3 years ago