Ask question

Ask question

All categories

2 years ago

5

You have the following problems and your friend gives you suggestions .work in pairs and have true conversation .Your prounction

is not so good.

1 answer:

Mrac [35]2 years ago

4 0

Answer:

hananzojsksnslapajzbzjsia

ja

Explanation:

kakaanzm

You might be interested in

A bullet is shot vertically upward with an initial velocity of 128 ft/s. The bullet's height after t seconds is y(t) = 128t - 16

saul85 [17]

The height of the bullet when the velocity is zero is 256 ft.

<h3>Height of the bullet when the velocity is zero </h3>

The height of the bullet when the velocity is zero is determined by taking derivative of the function as shown below;

$v = \frac{dy}{dt} = 128 -32t \\\\when \ v \ is \ zero\\\\v = 0\\\\128 - 32t = 0\\\\32t = 128\\\\t = \frac{128}{32} \\\\t = 4 \ s$

The height of the bullet at this time is calculated as follows;

$y(4) = 128(4) - 16(4)^2\\\\y(4) = 256 \ ft$

Learn more about height of projectiles here: brainly.com/question/10008919

6 0

2 years ago

A block of mass 0.500 kg is pushed against a horizontal spring of negligible mass until the spring is compressed a distance x (F

Marrrta [24]

Answer:

Part a)

$x = 0.4 m$

Part b)

$v_i = 11.7 m/s$

Part c)

Speed is more than the required speed so it will reach the top

Explanation:

Part a)

As we know that there is no frictional force while block is moving on horizontal plane

so we can use energy conservation on the block

$\frac{1}{2}mv^2 = \frac{1}{2}kx^2$

$\frac{1}{2}0.500(12^2) = \frac{1}{2}(450)x^2$

$x = 0.4 m$

Part b)

If the track has average frictional force of 7 N then work done by friction while block slides up is given as

$W_f = -7( \pi R)$

$W_f = -7(\pi \times 1.00)$

$W_f = -22 J$

work done against gravity is given as

$W_g = - mg(2R)$

$W_g = -(0.500)(9.8)(2\times 1)$

$W_g = -9.8 J$

Now by work energy equation we have

$\frac{1}{2}mv_i^2 + W_f + W_g = \frac{1}{2}mv_f^2$

$\frac{1}{2}0.5(12^2) - 9.8 - 22 = \frac{1}{2}(0.5)v_f^2$

$v_f = 4.1 m/s$

Part c)

now minimum speed required at the top is such that the normal force must be zero

$mg = \frac{mv^2}{R}$

$v = \sqrt{Rg}$

$v = 3.13 m/s$

so here we got speed more than the required speed so it will reach the top

5 0

3 years ago

a length 650 cm of thin thread wraps around a cylinder exactly 30 times calculate the circumference and the radius of the cylind

Eduardwww [97]

Answer:

circumference= 65/3 cm = 21.67 cm

radius R = 3.45 cm

Explanation:

To calculate the length of the circumference of the cylinder, we divide 650 cm by 30 (the number of times it wrapped exactly around it)

length of circumference= 65/3 cm = 21.67 cm

now use the formula of the circumference length to find the radius (R):

circumference length = 2 * pi * R

65/3 = 2 * pi * R

R = 65 / (6 pi)

R = 3.45 cm

5 0

3 years ago

The Surface Pressure at Leh, Ladakh is 800 mb. Now, assuming that Leh is at an altitude of 3500 m and every 100 m increase in he

Basile [38]

We have that the sea level pressure for Leh area is 1150mb mathematically given as

Ps= 1150 mb

<h3> Sea level pressure</h3>

Question Parameters:

Ladakh is 800 mb.

<u>assuming </u>that Leh is at an altitude of 3500 m and every 100 m

increase in height with respect to sea level corresponds to 10 mb pressure,

Generally, for 3500m the pressure change will be 350 mb.

Therefore, here for the sea level <em>pressure</em> we need to add,

Ps=800+350

Ps= 1150 mb

For more information on Pressure visit

brainly.com/question/25688500

8 0

2 years ago

skateboarder, starting from rest, rolls down a 13.5 m ramp. When she arrives at the bottom of the ramp her speed is 7.37 m/s. If

scZoUnD [109]

Answer:

1.7 m/s²

Explanation:

d = length of the ramp = 13.5 m

v₀ = initial speed of the skateboarder = 0 m/s

v = final speed of the skateboarder = 7.37 m/s

a = acceleration

Using the equation

v² = v₀² + 2 a d

7.37² = 0² + 2 a (13.5)

a = 2.01 m/s²

θ = angle of the incline relative to ground = 29.9

a' = Component of acceleration parallel to the ground

Component of acceleration parallel to the ground is given as

a' = a Cosθ

a' = 2.01 Cos29.9

a' = 1.7 m/s²

7 0

3 years ago