What metric units are used to report the mass of an object

A fast teaub abd a ski train driving from city a and city b face to face the fast train has a speed of 76km/h, the slow train ha

LekaFEV [45]

The distance between the two cities is 513.24 km.

<h3>Time of motion when the two trains meet</h3>

The time spent on the journey when the two trains meet is calculated as follows;

(Va - Vb)t = d

where;

d is the distance between the trains before meeting

(76 - 65)t = 40

11t = 40

t = 40/11

t = 3.64 hr

<h3>Distance traveled by the fast train</h3>

d1 = 76 km/h x 3.64 h

d1 = 276.64 km

<h3>Distance traveled by the slow train</h3>

d2 = 65 km/h x 3.64 h

d2 = 236.6 km

The distance between the two cities = 276.64 km + 236.6 km

= 513.24 km

Learn more about relative velocity here: brainly.com/question/17228388

7 0

2 years ago

A body is under the action of two forces 7N and 10N. Find the resultant of the two forces if the forces are parallel and act in

snow_lady [41]

Answer:K;O9KGVB

Explanation:

3 0

3 years ago

A 81 kg man is riding on a 40 kg cart traveling at a speed of 2.3 m/s. He jumps off with zero horizontal speed relative to the g

Alexus [3.1K]

Answer:

$\Delta v= 4.66\frac{m}{s}$

Explanation:

In this case we have to use the Principle of conservation of Momentum:

<em>This principle says that in a system the total momentum is constant if no external forces act in the system. The formula is:</em>

$m_1v_1+m_2v_2=m_1u_1+m_2u_2$

<em>Where:</em>

$m_1:$ Mass of the first object.

$m_2:$ Mass of the second object.

$v_1:$ Initial velocity of the first object.

$v_2:$ Initial velocity of the second object.

$u_1:$ Final velocity of the first object.

$u_2:$ Final velocity of the second object.

In <u>this problem</u> we have:

$m_1=81kg\\m_2=40kg\\v_1_2=2.3\frac{m}{s}$

$u_1=0\frac{m}{s}$

Observation: $v_1_2:$ Is because the system has the same initial velocity.

First we have to find $u_2$ ,

$m_1v_1+m_2v_2=m_1u_1+m_2u_2$

We can rewrite it as:

$(m_1+m_2)v_1_2=m_1u_1+m_2u_2$

Replacing with the data:

$(m_1+m_2)v_1_2=m_1u_1+m_2u_2\\\\(81kg+40kg)2.3\frac{m}{s}=81kg(0\frac{m}{s})+40kg(u_2)\\\\(121kg)2.3\frac{m}{s}=40kg(u_2)\\\\\frac{(121kg)2.3\frac{m}{s}}{40kg}=u_2\\\\\frac{278.3}{40}\frac{m}{s}=u_2\\\\6.96\frac{m}{s}=u_2$

We found the final velocity of the cart, but the problem asks for the resulting change in the cart speed, this means:

$\Delta v=u_2-v_2\\\Delta v=6.96\frac{m}{s}-2.3\frac{m}{s}\\\Delta v= 4.66\frac{m}{s}$

Then, the resulting change in the cart speed is:

$\Delta v= 4.66\frac{m}{s}$

5 0

3 years ago

You are working in a shoe test laboratory measuring the coefficients of friction for running shoes on a variety of surfaces. The

AnnZ [28]

Answer:

0.75

Explanation:

Since the static frictional force is the maximum force applied just before sliding, our frictional force, F is 300 N.

Since F = μN where μ = coefficient of static friction and N = normal force = 400 N (which is the downward force applied against the surface).

So, μ = F/N

= 300 N/400 N

= 3/4

= 0.75

So, the coefficient of static friction μ = 0.75

6 0

3 years ago

A 0.20-kg object is attached to the end of an ideal horizontal spring that has a spring constant of 120 N/m. The simple harmonic

miss Akunina [59]

Answer:

0.07756 m

Explanation:

Given mass of object =0.20 kg

spring constant = 120 n/m

maximum speed = 1.9 m/sec

We have to find the amplitude of the motion

We know that maximum speed of the object when it is in harmonic motion is given by $v_{max}=A\omega$ where A is amplitude and $\omega$ is angular velocity

Angular velocity is given by $\omega=\sqrt{\frac{k}{m}}$ where k is spring constant and m is mass

So $v_{max}=A\sqrt{\frac{k}{m}}$

$A=V_{max}\sqrt{\frac{m}{k}}=1.9\times \sqrt{\frac{0.2}{120}}=0.07756 \ m$

3 0

3 years ago

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