A car headed north at 15.0 m/s experiences an acceleration of 2.50 m/s2 for 3.00 s. What is the final velocity of the car?

A student drops two metallic objects into a 120-g steel con- tainer holding 150 g of water at 25°C. One object is a 200-g cube o

marissa [1.9K]

Answer:

The mass of the aluminum chunk is 258 g

Explanation:

Given;

mass of steel container = 120-g

mass of water = 150 g

initial temperature of water, = 25°C

mass of copper cube, $M_{cu}$ = 200 g

initial temperature of the copper cube, $T_c_u$ = 85°C

initial temperature of the aluminum chunk $T_A_l$ = 5.0°C

Neglecting heat loss, heat exchanged by the two metallic objects is the same since initial temperature is equal to final temperature of water.

$M_{Al}C_{Al} \delta T_{Al} = M_{cu}C_{cu} \delta T_{cu}$

where;

$C_{AL}$ is specific heat capacity of aluminum

$\delta T_{Al}$ is change in temperature of aluminum

$C_c_u$ is the specific heat capacity of copper

$\delta T_c_u$ is the change in temperature of copper

$M_{Al}C_{Al} \delta T_{Al} = M_{cu}C_{cu} \delta T_{cu} \\\\M_{Al} = \frac{M_{cu}C_{cu} \delta T_{cu}}{C_{Al} \delta T_{Al}} \\\\M_{Al} = \frac{0.2*387*60}{900*20} = 0.258 \ kg$

Therefore, the mass of the aluminum chunk is 258 g

7 0

3 years ago

A 55 kg person falling with a velocity of 0.6

siniylev [52]

Answer:

What are we supposed to find, if it is kinetic energy then this is the solution.

K.E=1/2mv^2

K.E= kinetic energy

M=mass

V=velocity

K.E =0.5*55*0.6^2

K.E=9.9J

Explanation:

3 0

2 years ago

A plane is flying east when it drops some supplies to a designated target below. The supplies land after falling for 10 seconds.

Diano4ka-milaya [45]

solution:

As Given plane is flying in east direction.

It throws back some supplies to designated target.

Time taken by the supply to reach the target =10 seconds

g = Acceleration due to gravity = - 9.8 m/s²[Taken negative as object is falling Downwards]

As we have to find distance from the ground to plane which is given by d.

d = $\frac{1}{2}\times g\times t^2$

= $\frac{1}{2}\times (9.8) \times(100) =50\times 9.8=490$ meters

Distance from the ground where supplies has to be land to plane = Option B =490 meters

6 0

3 years ago

Read 2 more answers

Describe what happens to chromosomes before mitosis.

Brums [2.3K]

Explanation:

Before mitosis, the chromosomes are copied. They then coil up, and each chromosome looks like a letter X in the nucleus of the cell. The chromosomes now consist of two sister chromatids. Mitosis separates these chromatids, so that each new cell has a copy of every chromosome

6 0

3 years ago

A 10-kg package drops from chute into a 25-kg cart with a velocity of 3 m/s. The cart is initially at rest and can roll freely w

amid [387]

Answer:

(a) the final velocity of the cart is 0.857 m/s

(b) the impulse experienced by the package is 21.43 kg.m/s

(c) the fraction of the initial energy lost is 0.71

Explanation:

Given;

mass of the package, m₁ = 10 kg

mass of the cart, m₂ = 25 kg

initial velocity of the package, u₁ = 3 m/s

initial velocity of the cart, u₂ = 0

let the final velocity of the cart = v

(a) Apply the principle of conservation of linear momentum to determine common final velocity for ineleastic collision;

m₁u₁ + m₂u₂ = v(m₁ + m₂)

10 x 3 + 25 x 0 = v(10 + 25)

30 = 35v

v = 30 / 35

v = 0.857 m/s

(b) the impulse experienced by the package;

The impulse = change in momentum of the package

J = ΔP = m₁v - m₁u₁

J = m₁(v - u₁)

J = 10(0.857 - 3)

J = -21.43 kg.m/s

the magnitude of the impulse experienced by the package = 21.43 kg.m/s

(c)

the initial kinetic energy of the package is calculated as;

$K.E_i = \frac{1}{2} mu_1^2\\\\K.E_i = \frac{1}{2} \times 10 \times (3)^2\\\\K.E_i = 45 \ J\\\\$

the final kinetic energy of the package;

$K.E_f = \frac{1}{2} (m_1 + m_2)v^2\\\\K.E_f = \frac{1}{2} \times (10 + 25) \times 0.857^2\\\\K.E_f = 12.85 \ J$

the fraction of the initial energy lost;

$= \frac{\Delta K.E}{K.E_i} = \frac{45 -12.85}{45} = 0.71$

7 0

3 years ago