How much work must be done on a 750 kg car to slow it down from 1.0 x 10 km/h to 2 50.0 km/h?

Explain how thermal energy (temperature) affects chemical changes.

Monica [59]

If bonds are broken, the energy is released, and if bonds are formed, energy is absorbed. During conversions from chemical energy to thermal energy, the energy stored in the chemical bonds are released and this energy causes surrounding molecules to move faster thus increasing the thermal energy of a substance.

8 0

3 years ago

A ball with a momentum of 16 kg.M/s strikes a ball at rest. What is the total momentum of both the balls after the collision.

melomori [17]

Answer:

Total momentum = 16 Kgm/s

Explanation:

Let the momentum of the two balls be A and B respectively.

Momentum A = 16 kgm/s

Momentum B = 0 kgm/s (since the ball is at rest).

Total momentum = A + B

Total momentum = 16 + 0

Total momentum = 16 Kgm/s

Momentum can be defined as the multiplication (product) of the mass possessed by an object and its velocity. Momentum is considered to be a vector quantity because it has both magnitude and direction.

Mathematically, momentum is given by the formula;

$Momentum = mass * velocity$

8 0

3 years ago

Read 2 more answers

Starting from zero, the electric current takes 2 seconds to reach half its maximum possible value in an RL circuit with a resist

Leno4ka [110]

Answer:

time=4s

Explanation:

we know that in a RL circuit with a resistance R, an inductance L and a battery of emf E, the current (i) will vary in following fashion

$i(t)=\frac{E}{R}(1-e^\frac{-t}{\frac{L}{R}})$ , where $i$ max= $\frac{E}{R}$

Given that, at i(2)= $\frac{imax}{2} =\frac{E}{2R}$

⇒ $\frac{E}{2R}=\frac{E}{R}(1-e^\frac{-2}{\frac{L}{R}})$

⇒ $\frac{1}{2}=1-e^\frac{-2}{\frac{L}{R}}$

⇒ $\frac{1}{2}=e^\frac{-2}{\frac{L}{R}}$

Applying logarithm on both sides,

⇒ $log(\frac{1}{2})=\frac{-2}{\frac{L}{R}}$

⇒ $log(2)=\frac{2}{\frac{L}{R}}$

⇒ $\frac{L}{R}=\frac{2}{log2}$

Now substitute $i(t)=\frac{3}{4}imax=\frac{3E}{4R}$

⇒ $\frac{3E}{4R}=\frac{E}{R}(1-e^\frac{-t}{\frac{L}{R}})$

⇒ $\frac{3}{4}=1-e^\frac{-t}{\frac{L}{R}}$

⇒ $\frac{1}{4}=e^\frac{-t}{\frac{L}{R}}$

Applying logarithm on both sides,

⇒ $log(\frac{1}{4})=\frac{-t}{\frac{L}{R}}$

⇒ $log(4)=\frac{t}{\frac{L}{R}}$

⇒ $t=log4\frac{L}{R}$

now subs. $\frac{L}{R}=\frac{2}{log2}$

⇒ $t=log4\frac{2}{log2}$

also $log4=log2^{2}=2log2$

⇒ $t=2log2\frac{2}{log2}$

⇒ $t=4$

5 0

3 years ago

A billiard ball (ball #1) moving at 5.00 m/s strikes a stationary ball (ball #2) of the same mass. after the collision, ball #1

Rashid [163]

If it is completely elastic, you can calculate the velocity of the second ball from the kinetic energy
v1 = velocity of #1 
v1' = velocity of #1 after collision 
v2' = velocity of #2 after collision. 

kinetic energy: v1^2 = v1' ^2 + v2' ^2 (1/2 and m cancel out) 
5^2 = 4.35^2 + v2' ^2 
v2 = 2.46 m/s <--- ANSWER

4 0

3 years ago

Read 2 more answers

Suppose we take a 1 m long uniform bar and support it at the 33 cm mark. Hanging a 0.15 kg mass on the short end of the beam res

MakcuM [25]

Answer:

The mass of the beam is 0.074 kg

Explanation:

Given;

length of the uniform bar, = 1m = 100 cm

Set up this system with the given mass and support;

0-----------------33cm-----------------------------------100cm

↓ Δ ↓

0.15kg m

Where;

m is mass of the uniform bar

Apply the principle of moment to determine the value of "m"

sum of anticlockwise moment = sum of clockwise moment

0.15kg(33 - 0) = m(100 - 33)

0.15(33) = m(67)

$m = \frac{0.15kg(33 \ cm)}{67 \ cm}\\\\m = 0.074 \ kg$

Therefore, the mass of the beam is 0.074 kg

8 0

3 years ago