The question will be on one of the comments on the answers you give just now take the points

A 0.02kg dart is loaded into a toy spring gun by compressing the spring 6cm. If the spring has a constant of 20 N/m, find… a. Th

katrin2010 [14]

Answer:

a) 0.036 J b) 0.036J c) 0.036 d) 1.9m/s e) 0.18 m

Explanation:

Mass of the dart = 0.02kg, the spring was compressed to 6cm

Work needed to compress the spring = 1/2*k*x ^2 where k is the force constant of the spring in N/m, x is the distance it was compressed in m

Work needed to compress the spring = 0.5 * 20* 0.06^2 since 6cm = 6 / 100 = 0.06 m

Work needed to compress the spring = 0.036J

b) the total energy stored in the spring = the work done to compress the spring = 0.036J

c) kinetic energy of the dart as it leaves the the spring = elastic potential energy stored in the spring = the work done in compressing the = 0.036J using the law of conservation of energy; energy is neither created nor destroyed but transformed from one form to another.

d) 1/2mv^2 = 0.036

mv^2 = 0.036*2

v^2 = 0.036*2 / 0.02 = 3.6

v = √3.6 = 1.897 approx 1.9m/s

e) kinetic energy of the dart = work done against gravity to get the body to height h

Work done against gravity = potential energy conserved at height = -mgh g is negative because the motion is upward while gravity acts downward

0.036 = 0.02 * 9.81 * h

0.036 / ( 0.02*9.81) = h

h = 0.18 m

6 0

3 years ago

In which region of the periodic table do the elements NOT normally react chemically with other elements?

horsena [70]

We would say that it would be metals.

6 0

2 years ago

Read 2 more answers

A car moves forward up a hill at 12 m/s with a uniform backward acceleration of 1.6 m/s2. What is its displacement after 6 s?

Romashka [77]

Answer:

The displacement of the car after 6s is 43.2 m

Explanation:

Given;

velocity of the car, v = 12 m/s

acceleration of the car, a = -1.6 m/s² (backward acceleration)

time of motion, t = 6 s

The displacement of the car after 6s is given by the following kinematic equation;

d = ut + ¹/₂at²

d = (12 x 6) + ¹/₂(-1.6)(6)²

d = 72 - 28.8

d = 43.2 m

Therefore, the displacement of the car after 6s is 43.2 m

6 0

2 years ago

(1 point) A frictionless spring with a 3-kg mass can be held stretched 1.6 meters beyond its natural length by a force of 90 new

Nonamiya [84]

Answer:

x(t)=0.337sin((5.929t)

Explanation:

A frictionless spring with a 3-kg mass can be held stretched 1.6 meters beyond its natural length by a force of 90 newtons. If the spring begins at its equilibrium position, but a push gives it an initial velocity of 2 m/sec, find the position of the mass after t seconds.

Solution. Let x(t) denote the position of the mass at time t. Then x satisfies the differential equation

$m \frac{d^{2}x}{dt^{2}} +kx=0$

Definition of parameters

m=mass 3kg

k=force constant

e=extension ,m

ω =angular frequency

k=90/1.6=56.25N/m

ω^2=k/m= 56.25/1.6

ω^2=35.15625

ω=5.929

General solution will be

$x(t)=c1cos(ωt)+c2Sin(ωt)$

$x(t)=c1cos(5.929t)+c2Sin(5.929t)$

differentiating x(t)

dx(t)=-5.929c1sin(5.929t)+5.929c2cos(5.929t)

when x(0)=0, gives c1=0

dx(t0)=2m/s gives c2=0.337

Therefore, the position of the mass after t seconds is

x(t)=0.337sin((5.929t)

6 0

2 years ago

Several forces act on an object at rest. It is known that the sum of the forces acting on the object is zero. Which statement is

MAVERICK [17]

Answer:

When net force of zero acting on a ball which is at rest , then the object center of mass will not accelerate , but the object may begin to rotate .

Explanation:

Here when there is an object where several forces are acing upon are zero then the center of mass will not accelerate because we know that

$A_{com}=\frac{F_{net}}{M}$

Where $A_{com} =$ acceleration of center of mass

$F_{net}$ = net force = 0

So the acceleration of center of mass will be zero

But the torque ,may not be zero as torque is product of individual force and perpendicular force .

Since if torque is not equal to zero then the object may begin to rotate

4 0

3 years ago