WILL MARK BRAIN A=154 B=145 C=26 D=206

Which of the following statements about sidewall markings is correct? a. The load index is given as a letter. b. The traction an

vagabundo [1.1K]

Answer: all the above options are correct.

Explanation:

In sidewall markings,the load index is given as a letter,traction and temperature ratings are based on the speed rating of the tire,the tire's recommended inflation pressure and load are indicated and the DOT code indicates when and where the tire was made.

8 0

3 years ago

Select the correct answer. A credit union is a business that is owned by people who have something in common, offers you a place

Svetradugi [14.3K]

Answer:

The primary purpose in furthering their goal of service is to encourage members to save money. Another purpose is to offer loans to members. In fact, credit unions have traditionally made loans to people of ordinary means. YOU ARE PART OWNER. Credit unions are owned and controlled by the people, or members, who use their services. Your vote counts. A volunteer board of directors is elected by members to manage a credit union.

A credit union is a nonprofit financial cooperative owned by and operated for the benefit of its members. It accepts deposits, makes loans, and provides other services.

5 0

2 years ago

A friend tosses a baseball out of his second floor window with initial velocity of 4.3m/s(42degrees below the horizontal). The b

schepotkina [342]

b)Determine your horisontal distance from window (ans. 1.5 m)
c)Calc the speed of ball as you catch it (ans: 8.2m/s)

I dont get what 42 m below the horizontal is, can someone give me direction on how to do this?

7 0

2 years ago

Read 2 more answers

Find the magnitude of the electric field due to a charged ring of radius "a" and total charge "Q", at a point on the ring axis a

34kurt

Answer:

E= $\frac{KQ}{2\sqrt 2a^2}$

Explanation:

We are given that

Charge on ring= Q

Radius of ring=a

We have to find the magnitude of electric filed on the axis at distance a from the ring's center.

We know that the electric field at distance x from the center of ring of radius R is given by

$E=\frac{kQx}{(R^2+x^2)^{\frac{3}{2}}}$

Substitute x=a and R=a

Then, we get

$E=\frac{KQa}{(a^2+a^2)^{\frac{3}{2}}}$

$E=\frac{KQa}{(2a^2)^{\frac{3}{2}}}$

$E=\frac{KQa}{2\sqrt 2a^3}$

$E=\frac{KQ}{2\sqrt 2a^2}$

Where K= $9\times 10^9 Nm^2/C^2$

Hence, the magnitude of the electric filed due to charged ring on the axis of ring at distance a from the ring's center= $\frac{KQ}{2\sqrt 2a^2}$

4 0

3 years ago

When the mass of the bottle is 0.125 kg, the KE is______ kg m2/s2.

DiKsa [7]

Answers are:
(1) KE = 1 kg m^2/s^2
(2) KE = 2 kg m^2/s^2
(3) KE = 3 kg m^2/s^2
(4) KE = 4 kg m^2/s^2

Explanation:

(1) Given mass = 0.125 kg
speed = 4 m/s

Since Kinetic energy = (1/2)*m*(v^2)

Plug in the values:
Hence:
KE = (1/2) * 0.125 * (16)
KE = 1 kg m^2/s^2

(2) Given mass = 0.250 kg
speed = 4 m/s

Since Kinetic energy = (1/2)*m*(v^2)

Plug in the values:
Hence:
KE = (1/2) * 0.250 * (16)
KE = 2 kg m^2/s^2

(3) Given mass = 0.375 kg
speed = 4 m/s

Since Kinetic energy = (1/2)*m*(v^2)

Plug in the values:
Hence:
KE = (1/2) * 0.375 * (16)
KE = 3 kg m^2/s^2

(4) Given mass = 0.500 kg
speed = 4 m/s

Since Kinetic energy = (1/2)*m*(v^2)

Plug in the values:
Hence:
KE = (1/2) * 0.5 * (16)
KE = 4 kg m^2/s^2

3 0

2 years ago