find the distance traveled in km when the average speed is 60 km per hour and the time taken is 3 and 1/2 hours

A girl is sitting in a sled sliding horizontally along some snow (there is friction present). The mass of the girl is 29.8 kg an

jonny [76]

Answer:

28.81 m

Explanation:

Ff = -123

m * a = -123

(29.8+10.3) * a = -123

a = -123/40.1 = -3.07

We know,

v^2 = u^2 + 2as

0^2 = 13.3^2 + 2*(-3.07)*s

s = 176.89/6.14 = 28.81

[ If there's a problem with the solution, pleaase let me know ]

6 0

3 years ago

A 50 Kg box sits at rest on a 30 degree ramp where the coef of static friction is 0.5773. If your push was directed at an angle

emmasim [6.3K]

Given data:

m= 50 Kg,

W= m×g = 50 × 9.81 = 490.5 N

ramp angle (α) = 30 degrees,

coefficient of friction (μs) = 0.5773,

Push at an angle (Θ) = 40 degrees,

Determine: Push to get box move up (P)=?

From the figure,

Resolving the forces along the plane

W sinα + μs.R = P cos Θ --------------------- (i)

Resolving the forces perpendicular to the inclined plane

W cosα = R+Psin Θ => R= W cosα - Psin Θ -------------- (ii)

Solving (i) and (ii) and keeping μs = tan Φ, Φ = Θ 

Pmin = W sin( α +Θ )

= W[ sin α.Cos Θ + cos α.sin Θ]

= 490.5 [ (sin 30.cos40) + (cos30.sin 40)]

= 460.9 N

Minimum push required to move the box up the ramp is 460.9 N

7 0

3 years ago

Two balls of clay, with masses M1 = 0.49 kg and M2 = 0.47 kg, are thrown at each other and stick when they collide. Mass 1 has a

malfutka [58]

Answer:

a) $p_i=1.568\hat{i}+0.752 \hat{j}$

b) $v_{fx}=1.668\ m.s^{-1}$

c) $v_{fy}=0.7999\ m.s^{-1}$

Explanation:

Given masses:

$m_1=0.49\ kg$

$m_2=0.47\ kg$

Velocity of mass 1, $v_1=3.2 \hat{i}\ m.s^{-1}$

Velocity of mass 2, $v_2=1.6 \hat{j}\ m.s^{-1}$

a)

Initial momentum:

$p_i=m_1.v_1+m_2.v_2$

$p_i=0.49\times 3.2 \hat{i}+0.47\times 1.6 \hat{j}$

$p_i=1.568\hat{i}+0.752 \hat{j}$

b)

magnitude of initial momentum:

$p_i=\sqrt{1.568^2+0.752 ^2}$

$p_i=1.739\ kg.m.s^{-1}$

From the conservation of momentum:

$p_f=p_i$

$m_f.v_f=1.739$

$v_f=\frac{1.739}{0.49+0.47}$

$v_f=1.85\ m.s^{-1}$ is the magnitude of final velocity.

Direction of final velocity will be in the direction of momentum:

$tan\theta=\frac{0.752 }{1.568}$

$\theta=25.62^{\circ}$

$\therefore v_{fx}=1.85\ cos25.62^{\circ}$

$v_{fx}=1.668\ m.s^{-1}$

c)

Vertical component of final velocity:

$v_{fy}=1.85\ sin 25.62^{\circ}$

$v_{fy}=0.7999\ m.s^{-1}$

6 0

3 years ago

In 1-2 sentences explain how you would determine the constant k for a spring.

Dmitrij [34]

To perform an experiment to determine the force constant of a spring, you will need a stand with a boss and clamp, a spiral spring, a meter rule and different weights.

The setup is arranged as shown in the image attached. The natural length of the spring is first recorded. Different weights are added to the spring one after the other and the extension is recorded.

The weight is now plotted on the vertical axis and the extension is plotted on the horizontal axis. The slope of the graph is the force constant of the spring.

Learn more: brainly.com/question/10991960

3 0

3 years ago

What is the effect of_on_?

evablogger [386]

skin cancer and it cause to lost transportating process

6 0

3 years ago

find the distance traveled in km when the average speed is 60 km per hour and the time taken is 3 and 1/2 hours​

find the distance traveled in km when the average speed is 60 km per hour and the time taken is 3 and 1/2 hours