Soils with a pH of 6.5 are considered slightly acidic true or false

A 2.0 kg block is released from rest at the top of a curved incline in the shape of a quarter of a circle of radius R = 3.0 m. T

Zigmanuir [339]

The block has maximum kinetic energy at the bottom of the curved incline. Since its radius is 3.0 m, this is also the block's starting height. Find the block's potential energy PE :

PE = m g h

PE = (2.0 kg) (9.8 m/s²) (3.0 m)

PE = 58.8 J

Energy is conserved throughout the block's descent, so that PE at the top of the curve is equal to kinetic energy KE at the bottom. Solve for the velocity v :

PE = KE

58.8 J = 1/2 m v ²

117.6 J = (2.0 kg) v ²

v = √((117.6 J) / (2.0 kg))

v ≈ 7.668 m/s ≈ 7.7 m/s

3 0

3 years ago

the gravitational force exerted on a baseball is 2.21 N down, a pitcher throws the ball horizontally with velocity 18.0 m/s by u

Ilia_Sergeevich [38]

To have a weight of 2.21N., the ball's mass is (2.21/9.8) = .226kg.
a) d = 1/2 (vt), = 1/2 (18 x .17), = 1.53m. 
b) Acceleration of the ball = (v/t), = 18/.17, = 105.88m/sec^2. 
f = (ma), = .226 x 105.88, = 23.92N.

8 0

3 years ago

Read 2 more answers

Just asking how everyone is doing cause of the virus

Bess [88]

Answer:

We are all fiine

Explanation:

How are you doing also

3 0

3 years ago

Read 2 more answers

An archer pulls the bowstring back for a distance of 0.470 m before releasing the arrow. The bow and string act like a spring wh

never [62]

Answer:

(a) 46.94 J.

(b) 55.95 m/s

Explanation:

(a)

Potential Energy: This is the energy of a body, due to its position. The S.I unit of potential energy is Joules (J).

The formula of potential energy in a stretched spring is

Ep = 1/2ke² .......................... Equation 1

Where Ep = potential energy of the spring, k = Force constant of the spring, e = extension or compression.

Given: k = 425 N/m, e = 0.47 m.

Substitute into equation 1

Ep = 1/2(425×0.47²)

Ep = 46.94 J.

(b)

at the instant When the arrow leaves the bow, the potential energy of the arrow is converted kinetic energy of the bow.

I.e,

Ep = 1/2mv² ............. Equation 2

Where m = mass of the arrow, v = velocity of the arrow.

make v the subject of the equation

v = √(2Ep/m)............. Equation 3

Given: Ep = 46.94 J, m = 0.03 Kg.

Substitute into equation 3

v = √(2×46.96/0.03)

v = √(93.92/0.03)

v = √(3130.67)

v = 55.95 m/s

5 0

3 years ago

What is the IMA of the following pulley system? 34567

Lynna [10]

Answer:

IMA of given system = $\frac{F_{r} }{F_{e} }$

Explanation:

The "Ideal Mechanical advantage" (IMA) of given pulley is $\frac{F_{r} }{F_{e} }$ .

Ideal Mechanical advantage of a system is defined by the ratio of achieved or output force to the implied force. In the pulley system above, output force is the resistant force denoted by $F_{r}$ . The input force is analogous or equivalent to the effort applied i.e. $F_{e}$ .

Hence by dividing these two forces we calculate the IMA of the above mentioned pulley system which is $\frac{F_{r} }{F_{e} }$ .
Its mathematical reference would be:

IMA = $\frac{F_{r} }{F_{e} }$

6 0

3 years ago