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Oduvanchick [21]
3 years ago
5

Two spheres of equal mass, A and B. are projected off the edge of a 2.0 m bench. Sphere A has a horizontal velocity of 5.0 m/s a

nd sphere B has a horizontal velocity of 2.5 m/s.
If both spheres leave the edge of the table at the same instant, sphere A will land where in relation to B?
A) Both Spheres will land at the same distance from the table.
B) Sphere A will go farther than Sphere B.
C) There isn't enough information to answer this question.
D) Sphere B will go farther than Sphere A.​
Physics
1 answer:
Ne4ueva [31]3 years ago
8 0

Answer:

the answer is B the second one

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A box with a mass of 40 kg sits at rest on a frictionless tile floor. with your foot, you apply a 20 N force in a horizontal dir
Grace [21]

Answer:

0.5 m/s²

Explanation:

according to Newton's second law, we are goven a relationship between force, mass and acceleration, with the formula:

F = m×a

F for force

m for mass

a for acceleration

we use the given data and get:

20 = 40×a

we find a=20/40=0.5m/s²

4 0
2 years ago
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A horizontal spring-mass system has low friction, spring stiffness 160 N/m, and mass 0.3 kg. The system is released with an init
anygoal [31]

Answer:

(a) 0.38 m

(b) 2.78 m/s

(c) 0.11 watt

Explanation:

mass, m = 0.3 kg

spring constant, K = 160 N/m

initial compression, d = 12 cm = 01.2 m

initial speed, u = 3 m/s

(a) Let the maximum stretch is y.

Use conservation of energy

Initial potential energy + initial kinetic energy = final potential energy

0.5 x K x d² + 0.5 x m x u² = 0.5 x K x y²

160 x 0.12 x 0.12 + 0.3 x 0.12 x 0.12 = 160 x y²

2.304 + 0.00432 = 160 y²

y = 0.38 m

y = 38 cm

(b) Let v is the maximum speed.

The speed is maximum when the stretch in the spring is zero, so by use of conservation of energy

Initial potential energy + initial kinetic energy = final kinetic energy

0.5 x K x d² + 0.5 x m x u² = 0.5 x m x v²

160 x 0.12 x 0.12 + 0.3 x 0.12 x 0.12 = 0.3 x v²

2.304 + 0.00432 = 0.3 v²

v = 2.78 m/s

(c) The time period of the spring mass system is given by

T=2\pi\sqrt{\frac{m}{K}}

T=2\pi\sqrt{\frac{0.3}{160}}

T = 0.272 second

Energy dissipated per cycle = 0.03 J

Power, P = 0.03 / 0.272 = 0.11 Watt

5 0
3 years ago
What time will the lunar eclipse happen central time?.
alexandr402 [8]
It was about 9:30 p.m. sorry if the answer is wrong
7 0
2 years ago
A block weighs 15 n and is suspended from a spring that is attached to the ceiling. the spring stretches by 0.075 m from its uns
Illusion [34]

We can salve the problem by using the formula:

F=kx

where F is the force applied, k is the spring constant and x is the stretching of the spring.


From the first situation we can calculate the spring constant, which is given by the ratio between the force applied and the stretching of the spring:

k=\frac{F}{x}=\frac{15 N}{0.075 m}=200 N/m


By using the value of the spring constant we calculated in the first step, we can calculate the new stretching of the spring when a force of 33 N is applied:

x=\frac{F}{k}=\frac{33 N}{200 N/m}=0.165 m

4 0
3 years ago
There is a 90 kg woman attached at the end of a bungee cord (k = 35 N/m) that is experiencing simple harmonic motion. How long d
Cloud [144]

Answer:

The value is T  =10.1 \ s

Explanation:

From the question we are told that

    The mass of the woman is  m  = 90 \  kg  

    The spring constant of the bungee cord is  k  =  35 \  N/  m

Generally the period of the oscillation (i,e time taken to complete on  cycle ) is mathematically represented as

            T  = 2 \pi *  \sqrt{ \frac{m}{k} }

=>      T  = 2 * 3.142  *  \sqrt{ \frac{90 }{ 35} }

=>      T  =10.1 \ s

3 0
3 years ago
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