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2 years ago

A 0.38 kg drinking glass is filled with a hot liquid. The liquid transfers 7032 J of energy to the glass. If the

Physics

1 answer:

kaheart [24]2 years ago

7 0

Answer:

841 J/kg.K

Explanation:

The computation of the specific hear of the glass is shown below:

As we know that

E= cmΔt

where

c denotes specific heat

m denotes 0.38 kg

Δt = temperature = 22k

E denotes energy = 7032 J

Now

7032 J = (0.38) (22) (c)

7032 J = 8.36 (c)

So C = 7032 J ÷ 8.36

= 841 J/kg.K

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A hawk is flying horizontally at 18.0 m/s in a straight line, 230 m above the ground. A mouse it has been carrying struggles fre

Lisa [10]

Answer:

a) vd = 47.88 m/s

b) θ = 80.9°

c) t = 6.8 s

Explanation:

In the situation of the problem, you can assume that the trajectory of the hawk and the trajectory of the mouse form a rectangle triangle.

One side of the triangle is the horizontal trajectory of the hawk after 2.00s of flight, the other side of the triangle is the distance traveled by the mouse when it is falling down. And the hypotenuse is the trajectory of the hawk when it is trying to recover the mouse.

(a) In order to calculate the diving speed of the hawk, you first calculate the hypotenuse of the triangle.

One side of the triangle is c1 = (18.0m/s)(2.0s) = 36m

The other side of the triangle is c2 = 230m - 3m = 227 m

Then, the hypotenuse is:

$h=\sqrt{(36m)^2+(227m)^2}=229.83m$ (1)

Next, it is necessary to calculate the falling down time of the mouse, this can be done by using the following formula:

$y=y_o+v_ot+\frac{1}{2}gt^2$ (2)

yo: initial height = 230m

vo: initial vertical speed of the mouse = 0m/s

g: gravitational acceleration = -9.8m/s^2

y: final height of the mouse = 3 m

You replace the values of the parameters in (2) and solve for t:

$3=230-4.9t^2\\\\t=\sqrt{\frac{227}{4.9}}=6.8s$

The hawk traveled during 2.00 second in the horizontal trajectory, hence, the hawk needed 6.8s - 2.0s = 4.8 s to travel the distance equivalent to the hypotenuse to catch the mouse.

You use the value of h and 4.8s to find the diving speed of the hawk:

$v_d=\frac{229.83m}{4.8s}=47.88\frac{m}{s}$

The diving speed of the Hawk is 47.88m/s

(b) The angle is given by:

$\theta=cos^{-1}(\frac{c_1}{h})=cos^{-1}(\frac{36m}{229.83m})=80.9 \°$

Then angle between the horizontal and the trajectory of the Hawk when it is descending is 80.9°

(c) The mouse is falling down during 6.8 s

4 0

3 years ago

Uma carga puntiforme de + 3,0uC é colocada em um ponto P de um campo elétrico gerado por uma partícula eletrizada com carga desc

expeople1 [14]

Responda:

1) E = 6 × 10 ^ 6NC ^ -1 2) Q = 6 × 10 ^ -5

Explicação:

Dado o seguinte:

Carga (q) = 3uC = 3 × 10 ^ -6C

Força elétrica (Fe) = 18N

Intensidade do campo elétrico (E) =?

Lembre-se:

Força elétrica (Fe) = carga (q) * Intensidade do campo elétrico (E)

Fe = qE; E = Fe / q

E = 18N / (3 × 10 ^ -6C)

E = 6N / 10 ^ -6C

E = 6 × 10 ^ 6NC ^ -1

Lembre-se:

E = kQ / r ^ 2

E = intensidade do campo elétrico

Q = carga de origem

r = distância de espera = 30cm = 30/100 = 0,3m

K = 9,0 × 10 ^ 9

6 × 10 ^ 6 = (9,0 × 10 ^ 9 * Q) / 0,3 ^ 2

9,0 × 10 ^ 9 * Q = 6 × 10 ^ 6 * 0,09

Q = 0,54 × 10 ^ 6 / 9,0 × 10 ^ 9

Q = 0,06 × 10 ^ (6-9)

Q = 0,06 × 10 ^ -3

Q = 6 × 10 ^ -5 = 60 × 10 ^ -6 = 60μC

7 0

3 years ago

Two electrons are passing 20.0 mm apart. What is the electric repulsive force that they exert on each other

vladimir1956 [14]

15.0 I’m pretty sure that’s the answer to your question

4 0

2 years ago

Read 2 more answers

To what potential should you charge a 2.0 μF capacitor to store 1.0 J of energy?

Bess [88]

E = (1/2)CV²
1 = (1/2)*(2*10⁻⁶)V²
10⁶ = V²
1000 = V

You should charge it to 1000 volts to store 1.0 J of energy.

6 0

3 years ago

Uma barra de ferro homogênea possui, a 20 °C, o comprimento de 6 m. Se o coeficiente de dilatação linear do ferro é 1,2 ∙ 10⁻⁵ °

Natali [406]

Answer:

25°C

Explanation:

Using the linear expansivity formula expressed as;

∝ = ΔL/lΔθ

∝ is coefficient of lineat expansion = 1.2 ∙ 10⁻⁵ °C⁻¹

ΔL is the change in length = 6.00036-6

ΔL = 0.00036m

l is the original length = 6m

Δθ is the change in temperature =θ₂-20

Substituting into the formula;

1.2 ∙ 10⁻⁵ °C⁻¹ = 0.00036/6(θ₂-20)

cross multiply

1.2 ∙ 10⁻⁵ * 6 = 0.00036/(θ₂-20)

7.2 ∙ 10⁻⁵= 0.00036/(θ₂-20)

0.00036 = 7.2 ∙ 10⁻⁵(θ₂-20)

0.00036 = 7.2 ∙ 10⁻⁵θ₂-144∙ 10⁻⁵

7.2 ∙ 10⁻⁵θ₂ = 0.00036+0.00144

7.2 ∙ 10⁻⁵θ₂ = 0.0018

θ₂ = 0.0018/0.000072

θ₂ = 25°C

Hence the temperature at which this bar must be acidic for its compression is 6,00036 m is 25°C

4 0

2 years ago