The force applied to a moving object was 100 N. The object moved 5 m. How much work was done on the object

A force of 10 N causes a spring to extend by 20 mm. Find: a) the spring constant of the spring in N/m

Mars2501 [29]

Answer:

formula used K=F/∆l

∆l is the elongation of the spring

F=10N
∆l=20mm===> 0.02m
K=10N divided 0.02m= 500N/m

6 0

3 years ago

Whats the difference between kinetic and potential energy?

damaskus [11]

Kinetic energy is energy in motion and potential energy is stored energy

5 0

3 years ago

A 0.25-kg ball attached to a string is rotating in a horizontal circle of radius 0.5 m. If the ball revolves twice every second,

Aloiza [94]

Answer:

19.74 N

Explanation:

mass of ball (m) = 0.25 kg

radius (r) = 0.5 m

time (t) = 2 revolutions per seconds = 1/2 = 0.5 second per revolution

find the tension in the string

tension (T) = $\frac{mv^{2} }{r}$

where velocity (v) = $\frac{2πr}{t}$

tension now becomes (T) = $\frac{m}{r} x (\frac{2πr}{t})^{2}$

tension (T) = $\frac{4π^{2}rm }{t^{2} }$

now substituting the values of mass (m), time (t) and radius (r) into the equation above we have

tension (T) = $\frac{4π^{2}x0.5x0.25}{0.5^{2} }$

tension (T) = $2π^{2}$ = $2x3.142^{2}$ = 19.74 N

4 0

3 years ago

Determine the moment of the force at AAA about point PPP. Use a vector analysis and express the result in Cartesian vector form.

Dimas [21]

Answer:

τ = 0

Explanation:

At the moment it is defined

τ = F x r

In tete case they give us the strength and position in Cartesian form, so it is easier to solve the determinant

τ = $\left[\begin{array}{ccc}i&j&k\\F_{x}&F_{y}&F_{z}\\x&y&z\end{array}\right]$

Let's apply this expression to the exercise

a) P = (-6 i ^ -3j ^ -6 k ^) m

F = (-6 i ^ -3j ^ -6k ^) 103 N

τ = $\left[\begin{array}{ccc}i&j&k\\-6&-3&-6\\-6&-3&-6\end{array}\right]$

τ = i ^ (3 6 - 3 6) + j ^ (6 6 -6 6) + k ^ (6 3 - 3 6)

τ = 0

b) P = 24i ^ + 8j ^ + 9k ^

F = 24i + 8j + 9k

τ = i ^ (72-72) + j ^ (216-216) + k ^ (24 8 - 8 24)

τ = 0

c) P = -6i + 6j-4k

F = -6i + 6j-4k

τ = i ^ (24-24) + j ^ (- 24 + 24) + k ^ (-36 + 36)

τ = 0

.d) P = 24i-8j + 9k

Let's change the sign of strength

F = -24i + 8j-9k

Tae = (I j k 24 -8 9 -24 8 -9)

Tae = i ^ (72 -72) + j ^ (- 216 + 216) + k ^ (192-192)

Tae = 0

8 0

3 years ago

I really need help with this!! Will give brainliest if you help T^T

slava [35]

Answer:

What inferences can you make about the melting points of the different substances and the motion of their particles, based on the data? (ignore that needed it here so i could see it better.)

Explanation:

Butter has a lower melting point than the cheese and the wax. The motion of the cheese were a little separated while the butter articles have more space in between. The wax had the closest particles.

I dont know if that makes sense?

5 0

3 years ago