Physics. I need help

Determine the minimum applied force p required to move wedge a to the right. the spring is compressed a distance of 175 mm. negl

BARSIC [14]

<span>b. The coefficient of static friction for all contacting surfaces is Î¼s=0.35. neglect friction at the rollers.

</span>

6 0

3 years ago

Which statement is true about acceleration?<br> who is an underated physicist?

wlad13 [49]

Answer: A vehicle's capacity to gain speed within a short time...

4 0

3 years ago

Which of the following cars have the most kinetic energy

faltersainse [42]

<h2>Hey There!</h2><h2>_____________________________________</h2><h2>Answer:</h2>

$\huge\boxed{OptionA}$

mass = 4 kg

velocity = 5 m/s^2

<h3 /><h3>Orange truck:</h3>

Mass= 2kg

Velocity = 7m/s^2

mass = 6 kg

velocity = 4m/s^2

<h3 /><h3>Green Car:</h3>

Mass = 8 kg

Velocity = 3 m/s^2

<h2>_____________________________________</h2><h2>SOLUTION:</h2>

By the equation of kinetic energy,

K.E = $\frac{1}{2} mv^2$

Where,

K.E is kinetic energy

m is mass

v is velocity

<h2>_____________________________________</h2><h3>Kinetic energy of Blue car:</h3>

Directly substitute the variables in the equation,

K.E = $\frac{1}{2}x4x5^2$

Simplify the equation,

K.E = 50 J

<h2>_____________________________________</h2><h3>Kinetic Energy of Silver Car:</h3>

Directly substitude the variable in the equation,

K.E = $\frac{1}{2}x6x4^2$

Simplify the equation,

K.E = 48J

<h2>_____________________________________</h2><h3>Kinetic Energy of Green Car:</h3><h3 />

Substitute the variables in the equation,

K.E = $\frac{1}{2}x8x3^2$

Simplify the Equation,

K.E = 36J

<h2>_____________________________________</h2><h3>Kinetic Energy of Orange Truck:</h3><h3 />

Substitute the variable,

K.E = $\frac{1}{2}x 2x7^2$

Simplify the equation,

K.E = 49J

As you can see that the highest value of kinetic energy is of Blue SUV thus it will be out answer.

<h2>_____________________________________</h2><h2>Best Regards,</h2><h2>'Borz'</h2><h3 /><h3 /><h3 /><h3 /><h2 /><h2 />

6 0

3 years ago

A ship sets sail from Rotterdam, The Netherlands, heading due north at 8.00 m/s relative to the water. The local ocean current i

Nuetrik [128]

Answer:

The velocity of the ship relative to the earth V = 9.05 $\frac{m}{s}$

Explanation:

The local ocean current is = 1.52 m/s

Direction $\theta$ = 40°

Velocity component in X - direction $V_{x}$ = 1.52 $\cos 40$ °

$V_{x}$ = 1.164 $\frac{m}{s}$

Velocity component in Y - direction $V_{y}$ = 8 + 1.52 $\sin 40$ °

$V_{y}$ = 8.97 $\frac{m}{s}$

The velocity of the ship relative to the earth

$V = \sqrt{V_{x}^{2} + V_{y} ^{2} }$

Put the values of $V_{x}$ and $V_{y}$ we get,

⇒ $V = \sqrt{1.164^{2} + 8.97 ^{2} }$

⇒ V = 9.05 $\frac{m}{s}$

This is the velocity of the ship relative to the earth.

7 0

3 years ago

A meter stick is balanced at the 50 cm mark. You tie a 20 N weight at the 20 cm mark. Where should a 30 N weight be placed so th

vivado [14]

Answer:

20 cm to the right of the center or 20+50 = 70 cm from the left side.

Explanation:

The length of meter stick is 1 m = 100 cm

Balance point on 50 cm

From the center the 20 N weight is 50-20 = 30 cm

Torque is obtained when force is multiplied with the distance

As the force is conserved we have

$\dfrac{20}{30}\times 30=20\ cm$

The distance will be 20 cm to the right of the center or 20+50 = 70 cm from the left side.

7 0

3 years ago

Physics. I need help​

Physics. I need help