Ask question

Ask question

All categories

3 years ago

15

A marble is rolling across a smooth 1.2 m tall table at a velocity of 3 m/s. How far from the edge of the table does it land? (S

HOW ALL OF YOUR WORK)

1 answer:

SashulF [63]3 years ago

4 0

Answer:

S = 1/2 g t^2 where t is the time to fall 1.2 m

t = (2 S / g)^1/2 = (2 * 1.2 / 9.8) = .495 s

Sy = Vy T = 3 m/sec * .495 sec = 1.48 m distance from edge of table

(Rotational speed has no effect since table is smooth)

You might be interested in

39. I also use it when I do the homework. (par 1 line 3).

Yuri [45]

Answer:

A

Explanation:

4 0

3 years ago

Read 2 more answers

1. °C= (°F-32) X 5/9. what will be the answer?

aleksley [76]

Answer:

F=9/5(°C)+32

Explanation:

Lets take an example

Take 25°C

$\\ \sf\longmapsto \dfrac{9}{5}(25)+32$

$\\ \sf\longmapsto 9(5)+32$

$\\ \sf\longmapsto 45+32$

$\\ \sf\longmapsto 77°F$

3 0

3 years ago

Z. A force that gives a 8-kg objet an acceleration of 1.6 m/s^2 would give a 2-kg object an

Paladinen [302]

Answer:

$\boxed {\boxed {\sf D.\ 6.4\ m/s^2}}$

Explanation:

We need to find the acceleration of the 2 kilogram object. Let's complete this in 2 steps.

<h3>1. Force of 1st Object </h3>

First, we can find the force of the first, 8 kilogram object.

According to Newton's Second Law of Motion, force is the product of mass and acceleration.

$F=m \times a$

The mass of the object is 8 kilograms and the acceleration is 1.6 meters per square second.

m= 8 kg
a= 1.6 m/s²

Substitute these values into the formula.

$F= 8 \ kg * 1.6 \ m/s^2$

Multiply.

$F= 12.8 \ kg*m/s^2$

<h3>2. Acceleration of the 2nd Object </h3>

Now, use the force we just calculated to complete the second part of the problem. We use the same formula:

$F= m \times a$

This time, we know the force is 12.8 kilograms meters per square second and the mass is 2 kilograms.

F= 12.8 kg *m/s²
m= 2 kg

Substitute the values into the formula.

$12.8 \ kg*m/s^2= 2 \ kg *a$

Since we are solving for the acceleration, we must isolate the variable (a). It is being multiplied by 2 kg. The inverse of multiplication is division. Divide both sides of the equation by 2 kg.

$\frac {12.8 \ kg*m/s^2}{2 \ kg}= \frac{2\ kg* a}{2 \ kg}$

$\frac {12.8 \ kg*m/s^2}{2 \ kg}=a$

The units of kilograms cancel.

$\frac {12.8}{2}\ m/s^2=a$

$6.4 \ m/s^2=a$

The acceleration is 6.4 meters per square second.

4 0

2 years ago

A suspension bridge has twin towers that are 1300 feet apart. Each tower extends 180 feet above the road surface. The cables are

zhannawk [14.2K]

Answer:

$y = 17.04 ft$

Explanation:

Since the cable touches the road at the mid point of two towers

so here we have vertex at that mid point taken to be origin

now the maximum height on the either side is given as

$y = 180 ft$

horizontal distance of the tower from mid point is given as

$x = \frac{1300}{2} = 650 ft$

now from the equation of parabola we have

$y = k x^2$

$180 = k(650^2)$

$k = 4.26 \times 10^{-4}$

now we have

$y = (4.26 \times 10^{-4})x^2$

now we need to find the height at distance of 200 ft from center

so we have

$y = (4.26 \times 10^{-4})(200^2)$

$y = 17.04 ft$

8 0

3 years ago

Pls do question 1 part d). tysm

Aneli [31]

Answer:

c

Explanation:

cuz its informing the length of 5 and weight on 20N

5 0

3 years ago