1:04 here dims its too short write at least 20 characters

A car traveling at 30 m/s runs out of gas while traveling up a 10 slope. how far up the hill what coast before starting to roll

kaheart [24]

Answer:

264.7 m

Explanation:

The net force felt by the car travelling along the slope is equal to the component of the weight parallel to the slope, so:

$F=-mg sin \theta$

where

m is the mass of the car

g = 9.8 m/s^2 is the acceleration of gravity

$\theta=10^{\circ}$

and the negative sign is due to the fact the force is opposite direction to the motion of the car

The acceleration of the car is therefore:

$a=\frac{F}{m}=-gsin \theta=-9.8 sin 10^{\circ}=-1.7 m/s^2$

Now we can find how far the car went up the hill by using the equation:

$v^2-u^2 = 2ad$

where

v = 0 is the final velocity of the car

u = 30 m/s is the initial velocity

g = -1.7 m/s^2 is the acceleration

d is the distance covered

Solving for d,

$d=\frac{v^2-u^2}{2a}=\frac{0-30^2}{2(-1.7)}=264.7 m$

3 0

4 years ago

A 100 kg box is suspended from two ropes. The "left rope makes an angle of 20" degrees with the vertical, and the right rope mak

GarryVolchara [31]

Explanation:

It is given that,

Mass of the box, m = 100 kg

Left rope makes an angle of 20 degrees with the vertical, and the right rope makes an angle of 40 degrees.

From the attached figure, the x and y component of forces is given by :

$T_{1x} =-T_1 cos (20)$

$T_{2x} = T_2 cos (40)$

$mg_x = 0$

$T_{1y} = T_1 sin (20)$

$T_{2y} = T_2 sin (40)$

$mg_y= -mg$

Let $R_x$ and $R_y$ is the resultant in x and y direction.

$R_x=-T_1 cos (20)+T_2 cos (40)+0$

$R_y=T_1 sin(20)+T_2 sin(40)-mg$

As the system is balanced the net force acting on it is 0. So,

$-T_1 cos (20)+T_2 cos (40)+0=0$ .............(1)

$T_1 sin(20)+T_2 sin(40)-100\times 9.8=0$ ..................(2)

On solving equation (1) and (2) we get:

$T_1=866.86\ N$ (tension on the left rope)

$T_2=1063.36\ N$ (tension on the right rope)

So, the tension on the right rope is 1063.36 N. Hence, this is the required solution.

7 0

3 years ago

A 0.439 kg puck, initially at rest on a horizontal, frictionless surface, is struck by a 0.307 kg puck moving initially along th

lesya [120]

Answer:

u2 = 0.266 m/s

Explanation:

Let the left Puck mass at rest = m1 =0.307 Kg

mass of the right puck m2 = 0.439 kg

velocity of m1 before collision v1= 2.19 m/s

velocity of m2 before collision v2 = 0m/s

velocity of m1 after collision u1 =1.19 m/s

velocity of m2 after collision u2 = ? m/s

θ = 37°

Solution:

Before collision:

Momentum (y-axis ) before collision= 0 Kgm/s

Momentum (x-axis ) before collision= m1v1 + m2v2 = 0.307 Kg x 2.19 m/s + 0

= 0.672 Kgm/s

After collision:

Momentum (y-axis ) after collision= m1u1 sinθ + m2u2 sinθ

= 0.307 x 1.19 m/s sin 37 ° + 0.439 x u2 sin 37°

= 0.22 + 0.26 u2

Momentum (x-axis ) after collision= m1u1 cosθ + m2u2 cos θ

= 0.307 x 1.19 m/s cos 37 ° + 0.439 x u2 cos 37°

= 0.29 + 0.35 u2

According to law of conservation momentum

momentum before collision = momentum after collision.

0 + 0.672 Kgm/s = 0.22 Kgm/s + 0.26 kg u2 + 0.29 Kgm/s + 0.35 kg u2

0.672 Kgm/s = 0.51 Kgm/s + 0.61 u2

u2 = 0.266 m/s

6 0

4 years ago

Please helo me to 1st question

k0ka [10]

Explanation:

6 min in seconds:

6×60

<h2>360 seconds </h2><h2 /><h2 />

6 min in hours:

<h2>0.1 hours</h2>

6 0

3 years ago

Mechanical energy that has been ‘lost' to friction isn't really lost. It just is no longer in its mechanical form. True or False

Vadim26 [7]

Answer:

True.

Explanation:

Energy can be defined as the ability (capacity) to do work. The two (2) main types of energy are;

a. Gravitational potential energy (GPE): it is an energy possessed by an object or body due to its position above the earth.

b. Kinetic energy (KE): it is an energy possessed by an object or body due to its motion.

Furthermore, the mechanical energy of a physical object or body is the sum of the potential energy and kinetic energy possessed by the object or body.

Mathematically, it is given by the formula;

Mechanical energy = G.P.E + K.E

Mechanical energy that has been ‘lost' to friction isn't really lost. It just is no longer in its mechanical form. This is ultimately in accordance with the law of conservation of energy, which states that energy cannot be destroyed but can only be converted or transformed from one form to another.

Hence, Mechanical energy that has been ‘lost' to friction isn't really lost but converted into heat energy.

4 0

3 years ago