All categories

3 years ago

Read this because I need help pleaseeeeee !!!

Physics

1 answer:

larisa86 [58]3 years ago

4 0

There are several formulas that describe the distance, speed, and time for
a falling body. The one I use the most happens to be the one that'll be the
most useful to solve this problem:

      Distance = 1/2 g t²

We know the distance and we know ' g ', so we can use
this formula to find ' t '.

   Distance = (1/2) (gravity) (time)²

   (239 m) = (1/2) (3.7 m/s²) (time)²

Divide each side
by 1.85 m/s² : (129 m) / (1.85 m/s²) = (time)²

   (129/1.85) sec² = (time)²

Take the square root
of each side: 8.35 sec = time

You might be interested in

Marina is staring at an optical illusion where he sees a version of the American flag that is colored green, yellow, and black.

k0ka [10]

Explanation:

The reason why Marina sees the colour red, white and blue or the original colour of the American flag is that because of a phenomenon known as Afterimage. The retina in our eyes have mainly three receptors that are colour sensitive known as cones. These receptors can perceive the colour green, red and blue. Now when we look or stare at a particular colour for a long time, what happen is that our retina becomes tired and they ignore the colours that stared at. And now they work to form other colours at the retina just like the way when we produce other colour from the primary colour.

If the red receptor gets exhausted we will see the colour red. Likewise when we see the colour orange when we stare at the colour blue.

This explains the optical illusion of the American flag.

6 0

3 years ago

A ladder rests against a vertical wall at a point 12 feet from the floor. The angle formed by the ladder and the floor is 63°. C

GenaCL600 [577]

Answer:

length of the ladder is 13.47 feet

base of wall to latter distance 6.10 feet

angle between ladder and the wall is 26.95°

Explanation:

given data

height h = 12 feet

angle 63°

to find out

length of the ladder ( L) and length of wall to ladder ( A) and angle between ladder and the wall

solution

we consider here angle between base of wall and floor is right angle

we apply here trigonometry rule that is

sin63 = h/L

put here value

L = 12 / sin63

L = 13.47

so length of the ladder is 13.47 feet

and

we can say

tan 63 = h / A

put here value

A = 12 / tan63

A = 6.10

so base of wall to latter distance 6.10 feet

and

we say here

tanθ = 6.10 / 12

θ = 26.95°

so angle between ladder and the wall is 26.95°

8 0

3 years ago

Friends Burt and Ernie stand at opposite ends of a uniform log that is floating in a lake. The log is 3.0 m long and has mass 20

Taya2010 [7]

Answer:

The distance the log has moved by the time Ernie reaches Bur is 1.33 m.

Explanation:

give information:

The log is 3.0 m long and has mass 20.0 kg.

Burt has mass 30.0 kg; Ernie has mass 40.0 kg

Ernie has mass 40.0 kg.

to find the distance, first, we have to calculate the center of mass

X = ∑ m x /∑m

= (20 x (3/2)) + (30 x 0) + (40 x 3)/ (20+30+40)

= 150/90

= 5/3

when Ernie walk, the center of the mass is

X = (70 x 0) + (20 x (3/2))/(70 + 20)

= 30/90

= 1/3

the distance of log moved = 5/3 - 4/3 = 1.33 m

5 0

3 years ago

The slope of a distance-time graph will give

Dafna1 [17]

5.
explanation: the answer is 5

5 0

3 years ago

A hanging weight, with a mass of m1 = 0.365 kg, is attached by a string to a block with mass m2 = 0.825 kg as shown in the figur

morpeh [17]

The speed of the block after it has moved the given distance away from the initial position is 1.1 m/s.

<h3>Angular Speed of the pulley </h3>

The angular speed of the pulley after the block m1 fall through a distance, d, is obatined from conservation of energy and it is given as;

K.E = P.E

$\frac{1}{2} mv^2 + \frac{1}{2} I\omega^2 = mgh\\\\\frac{1}{2} m_2v_0^2 + \frac{1}{2} \omega^2(m_1R^2_2 + m_2R_2^2) + \frac{1}{2} \omega^2( \frac{1}{2} MR_1^2 + \frac{1}{2} MR_2^2) = m_1gd- \mu_km_2gd\\\\\frac{1}{2} m_2v_0^2 + \frac{1}{2} \omega^2[R_2^2(m_1 + m_2)+ \frac{1}{2} M(R_1^2 + R_2^2)] = gd(m_1 - \mu_k m_2)\\\\$

$\frac{1}{2} m_2v_0 + \frac{1}{4} \omega^2[2R_2^2(m_1 + m_2) + M(R^2_1 + R^2_2)] = gd(m_1 - \mu_k m_2)\\\\2m_2v_0 + \omega^2 [2R_2^2(m_1 + m_2) + M(R^2_1 + R^2_2)] = 4gd(m_1 - \mu_k m_2)\\\\\omega^2 [2R_2^2(m_1 + m_2) + M(R^2_1 + R^2_2)] = 4gd(m_1 - \mu_k m_2) - 2m_2v_0^2\\\\\omega^2 = \frac{ 4gd(m_1 - \mu_k m_2) - 2m_2v_0^2}{2R_2^2(m_1 + m_2) + M(R^2_1 + R^2_2)} \\\\\omega = \sqrt{\frac{ 4gd(m_1 - \mu_k m_2) - 2m_2v_0^2}{2R_2^2(m_1 + m_2) + M(R^2_1 + R^2_2)}} \\\\$

Substitute the given parameters and solve for the angular speed;

$\omega = \sqrt{\frac{ 4(9.8)(0.7)(0.365 \ - \ 0.25\times 0.825) - 2(0.825)(0.82)^2}{2(0.03)^2(0.365 \ + \ 0.825)\ \ +\ \ 0.35(0.02^2\ + \ 0.03^2)}} \\\\\omega = \sqrt{\frac{3.25}{0.00214\ + \ 0.000455 } } \\\\\omega = 35.39 \ rad/s$

<h3>Linear speed of the block</h3>

The linear speed of the block after travelling 0.7 m;

v = ωR₂

v = 35.39 x 0.03

v = 1.1 m/s

Thus, the speed of the block after it has moved the given distance away from the initial position is 1.1 m/s.

Learn more about conservation of energy here: brainly.com/question/24772394

5 0

2 years ago