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4 years ago

10. A cat walking at 0.25 m/s sees a mouse and accelerates uniformly at 0.40 m/s

Physics

1 answer:

NNADVOKAT [17]4 years ago

3 0

Answer:

The displcament is 2.55m

Explanation:

step one:

Given that the intial speed is

u= 0.25 m/s

and acceleration is

a= 0.40 m/s^2

and time t= 3seconds

step two:

Applying the formula

s=ut+1/2at^2

s= 0.25*3+ 1/2*0.4*3^3

s= 0.75+3.6/2

s= 0.75+1.8

s=2.55 m

The displacement is 2.55m

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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 3500 N is force is applied to a spring that has a spring of constant of k= 14000 N/m. How far from equilibrium will the spring

Taya2010 [7]

Answer:

the spring be displaced by 25.0 cm

Explanation:

The computation is shown below:

As we know that

F= -K × x

So,

$x = \frac{-F}{K}$

Now

$x = \frac{-3500}{14000} \\\\$

= -0.250m

= 25.0 cm

Hence, the spring be displaced by 25.0 cm

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3 years ago

In this type of bond, electrons are lost or gained by atoms, and the atoms are held together by electrical attraction.

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4 years ago

A stone is dropped into a river from a bridge at a height h above the water. Another stone is thrown vertically down at a time t

Mumz [18]

Answer:

$v_{y_0} = \frac{\frac{g}{2}t(t - 2\sqrt{\frac{2h}{g}})}{\sqrt{\frac{2h}{g}} - t}$

Explanation:

We will apply the equations of kinematics to both stones separately.

First stone:

Let us denote the time spent after the second stone is thrown as 'T'.

$y - y_0 = v_{y_0}(t+T) + \frac{1}{2}a(t+T)^2\\0 - h = 0 + \frac{1}{2}(-g)(t+T)^2\\(t+T)^2 = \frac{2h}{g}\\T = \sqrt{\frac{2h}{g}}-t$

Second stone:

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