What are the size and direction of the force that is the sum of a force of 3 n acting south and a force of 6 n acting north?

A 1.0-kg block of aluminum is at a temperature of 50 Celsius. How much thermal energy will it lose when it’s temperature is redu

Ipatiy [6.2K]

Answer:

The lose of thermal energy is, Q = 22500 J

Explanation:

Given data,

The mass of aluminium block, m = 1.0 kg

The initial temperature of block, T = 50° C

The final temperature of the block, T' = 25° C

The change in temperature, ΔT = 50° C - 25° C

= 25° C

The specific heat capacity of aluminium, c = 900 J/kg°C

The formula for thermal energy,

Q = mcΔT

= 1.0 x 900 x 25

= 22500 J

Hence, the lose of thermal energy is, Q = 22500 J

7 0

3 years ago

Find the voltage across the 15 Ω resistor. . 35 Ω 20 Ω 15 Ω 10 V

almond37 [142]

Answer:

35 Ω

Explanation:

3 0

2 years ago

Which of the following is the smallest conceivable amount of time that could pass between a lunar eclipse and a solar eclipse .

pickupchik [31]

The time required for a moon to orbit around the earth is about 27-28 days

In order for lunar eclipse to occur the line that should be formed is:
Sun-Earth-Moon
because earth is making shade on moon

in order for solar eclipse to occur the line is now:
Sun-Moon-Earth
because moon is making a shade on earth (blocking sun = solar eclipse)

Therefore moon needs to make half of its orbit to go from behind the earth to in front of the earth.

28/2 = 14

Answer is 14

5 0

2 years ago

What happens to its kinetic energy when its reaches the maximum height ?

drek231 [11]

when it reaches the maximum height, all the energy has now been converted into potential energy.when a ball is thrown straight upto into the air,all its initial kinetic energy converted into gravitational potential energy when it reaches its maximum height

7 0

3 years ago

Two blocks are connected by a light weight, flexible cord that passes over a frictionless pulley.Ifm1=2 kg and m2 = 3 kg, and bl

Vera_Pavlovna [14]

Answer:

t = 1.41 sec.

Explanation:

If we assume that the acceleration of the blocks is constant, we can apply any of the kinematic equations to get the time since the block 2 was released till it reached the floor.

First, we need to find the value of acceleration, which is the same for both blocks.

If we take as our system both blocks, and think about the pulley as redirecting the force simply (as tension in the strings behave like internal forces) , we can apply Newton's 2nd Law, as they were moving along the same axis, aiming at opposite directions, as follows:

F = m₂*g - m₁*g = (m₁+m₂)*a (we choose as positive the direction of the acceleration, will be the one defined by the larger mass, in this case m₂)

⇒ a = ( $\frac{(m₂-m₁)}({m₁+m₂} * g$ = g/5 m/s²

Once we got the value of a, we can use for instance this kinematic equation, and solve for t:

Δx = 1/2*a*t² ⇒ t² = (2* 1.96m *5)/g = 2 sec² ⇒ t = √2 = 1.41 sec.

6 0

3 years ago