All categories

3 years ago

A baseball rolls off of a 0.70 m high desk at 0.66m/s.How far from the base of the desk will it hit the floor?

Physics

1 answer:

sergij07 [2.7K]3 years ago

5 0

0.66 m/s is the answer to this question final answer

You might be interested in

Suppose you lift a 20 kg box by a height of 1.0 m.

9966 [12]

W=mgh W=(20)(9.8)(1) w=196J

6 0

3 years ago

Read 2 more answers

A 0.50-kg mass attached to the end of a string swings in a vertical circle (radius 2.0 m). When the mass is at the highest point

il63 [147K]

Answer:

31.1 N

Explanation:

m = mass attached to string = 0.50 kg

r = radius of the vertical circle = 2.0 m

v = speed of the mass at the highest point = 12 m/s

T = force of the string on the mass attached.

At the highest point, force equation is given as

$T + mg =\frac{mv^{2}}{r}$

Inserting the values

$T + (0.50)(9.8) =\frac{(0.50)(12)^{2}}{2}$

T = 31.1 N

7 0

3 years ago

Read 2 more answers

Why do we tend to attribute others' negative behaviors to dispositions over situational factors?

miskamm [114]

Answer:

Because you have a greater understanding of the mentality and conduct of individuals you care about, you are better able to understand their perspective and are more likely to recognize possible situational factors for their behavior.

4 0

2 years ago

As an aid in working this problem, consult Interactive Solution 3.41. A soccer player kicks the ball toward a goal that is 20.0

alekssr [168]

Answer:

V=14.9 m/s

Explanation:

In order to solve this problem, we are going to use the formulas of parabolic motion.

The velocity X-component of the ball is given by:

$Vx=V*cos(\alpha)\\Vx=15.7*cos(31^o)=13.5m/s$

The motion on the X axis is a constant velocity motion so:

$t=\frac{d}{Vx}\\t=\frac{20.0}{13.5}=1.48s$

The whole trajectory of the ball takes 1.48 seconds

We know that:

$Vy=Voy+(a)*t\\Vy=15.7*sin(31^o)+(-9.8)*(1.48)=-6.42m/s$

Knowing the X and Y components of the velocity, we can calculate its magnitude by:

$V=\sqrt{Vx^2+Vy^2} \\V=\sqrt{(13.5)^2+(-6.42)^2}=14.9m/s$

6 0

3 years ago

Student 1 lifts a box with a force of 500 N and sets it on a tabletop 1.2 m high. Student 2 pushes an identical box up a 5 m ram

Troyanec [42]

The student who did the most work is student 2 with 2500 Joules.

Given the following data:

Force 1 = 500 Newton

Distance 1 = 1.2 meter

Force 2 = 500 Newton

Distance 2 = 5 meter

To determine which of the students did the most work:

Mathematically, the work done by an object is given by the formula;

$Work\;done = Force \times distance$

For student 1:

$Work\;done = 500 \times 1.2$

Work done = 600 Joules

For student 2:

$Work\;done = 500 \times 5$

Work done = 2500 Joules.

Therefore, the student who did the most work is student 2 with 2500 Joules.

Read more: Read more: brainly.com/question/13818347

7 0

3 years ago

Read 2 more answers