Ask question

Ask question

All categories

gizmo_the_mogwai [7]

3 years ago

5

A train travels 65 kilometers in 5 hours, and then 61 kilometers in 4 hours. What is its average speed? _______km/hr

2 answers:

olasank [31]3 years ago

4 0

The trains average speed is 13 km/hr.

Galina-37 [17]3 years ago

4 0

Yes, Average speed equals total distance divided by total time.

Total distance= 65km + 61km= 126 km

Total time= 5h + 4h = 9 h

126km/9h= 14km/h

The trains average speed is 14 km/h

You might be interested in

A stone is dropped from a cliff. after it has fallen 10m what is the stones velocity

Ludmilka [50]

The answer is 14,1421 m/s. :)

3 0

3 years ago

Read 2 more answers

The activity of a radioisotope is found to decrease 40% of its original value in 2.59 x 10 s.

Rainbow [258]

Answer: $0.0353\ s^{-1}$

Explanation:

Given

Radioactive material is found to decrease 40% of its original value in $2.59\times 10\ s$

Sample at any time is given by

$N=N_oe^{-\lambda t}$

where, $\lambda=\text{decay constant}$

Put values

$\Rightarrow 0.4N_o=N_oe^{-\lambda\cdot 2.59\times 10}\\\Rightarrow 0.4=e^{-\lambda\cdot 2.59\times 10$

Taking natural logarithm both side

$\Rightarrow \lambda=\dfrac{\ln 2.5}{25.9}\\\\\Rightarrow \lambda =0.0353\ s^{-1}$

8 0

3 years ago

beks73 [17]

Answer:

2 m/s²

Explanation:

the equations of motion are

S= ut +½at²

v² = u²+ 2as

v = u + at

s = (u+v)/2 × t

From the parameters given

u = 0m/s this is because it starts from rest

Distance (s) = 9m

Time (t) = 3s

Based on this the first equation would be used

s = ut + ½at²

Input values

9 = 0×3 + ½ × a x 3²

9 = 0 + 9a/2

9 = 4.5a

Divide both sides by 4.5

a = 9 / 4.5 m/s²

a = 2 m/s²

I hope this was helpful, please mark as brainliest

3 0

3 years ago

A child in danger of drowning in a river is being carried downstream by a current that flows due south uniformly with a speed of

tia_tia [17]

Let the rescue boat starts at an angle theta with the North

now its velocity towards East is given as

$v_x = 24sin\theta$

$v_y = -24cos\theta + 3$

now in some time "t" it will catch the boy

so we will have

$t = \frac{0.5}{24sin\theta}$

also we have

$t = \frac{2}{-24cos\theta + 3}$

now we have

$\frac{2}{-24cos\theta + 3} = \frac{0.5}{24sin\theta}$

$4*24sin\theta = - 24cos\theta + 3$

$96 sin\theta + 24cos\theta = 3$

by solving above we got

$\theta = 164 degree$

3 0

3 years ago

A student sits on a rotating stool holding two 3.09-kg masses. When his arms are extended horizontally, the masses are 1.08 m fr

schepotkina [342]

Answer:

a

The New angular speed is $w_f = 2.034 rad/s$

b

The Kinetic energy before the masses are pulled in is $KE_i = 3.101 \ J$

c

The Kinetic energy after the masses are pulled in is $KE_f = 8.192 \ J$

Explanation:

From the we are told that masses are 1.08 m from the axis of rotation, this means that

The radius $r =1.08m$

The mass is $m = 3.09\ kg$

The angular speed $w = 0.770 \ rad/sec$

The moment of inertia of the system excluding the two mass $I = 3.25 \ kg \cdot m^2$

New radius $r_{new} = 0.34m$

Generally the conservation of angular momentum can be mathematical represented as

$w_f = [\frac{I_i}{I_f} ]w_i .....(1)$

Where $w_f$ is the final angular speed

$w_i$ is the initial angular speed

$I_i$ is the initial moment of inertia

$I_f$ is the final moment of inertia

Moment of inertia is mathematically represented as

$I = m r^2$

Where I is the moment of inertia

m is the mass

r is the radius

So the Initial moment of inertia is given as

$I_i = moment \ of \ inertia \ of\ the \ two \ mass \ + 3.25 \ kg \cdot m^2$

$I_i = 2m r^2 + 3.25$

The multiplication by is because we are considering two masses

$I_i = 2 [(3.09)(1.08)^2] +3.25 = 10.46 kg \cdot m^2$

So the final moment of inertia is given as

$I_f = 2[(3.09)(0.34)^2] +3.25 = 3.96 \ kg \cdot m^2$

Substituting these values into equation 1

$w_f = [\frac{10.46}{3.96} ] * 0.77 = 2.034 \ rad/sec$

Generally Kinetic energy is mathematically represented in term of moment of inertia as

$KE = \frac{1}{2} * I * w^2$

Now considering the kinetic energy before the masses are pulled in,

$KE_i = \frac{1}{2} * I_i * w^2_i$

The Moment of inertia would be $I_i = 10.46 \ Kg \cdot m^2$

The Angular speed would be $w_i = 0.77 \ rad/s$

Now substituting these value into the equation above

$KE_i = \frac{1}{2} * (10.46) * (0.770)^2 = 3.101 J$

Now considering the kinetic energy after the masses are pulled in,

$KE_f = \frac{1}{2} * I_f * w^2_f$

The Moment of inertia would be $I_f = 3.96 \ Kg \cdot m^2$

The Angular speed would be $w_f = 2.034 \ rad/s$

Now substituting these value into the equation above

$KE_f= \frac{1}{2} *(3.96)(2.034)^2$

$= 8.192J$

8 0

3 years ago