Answer:
17 °C
Explanation:
From specific Heat capacity.
Q = cm(t₂-t₁)................. Equation 1
Where Q = Heat absorb by the metal block, c = specific heat capacity of the metal block, m = mass of the metal block, t₂ = final temperature, t₁ = Initial temperature.
make t₁ the subject of the equation
t₁ = t₂-(Q/cm)............... Equation 2
Given: t₂ = 22 °C, Q = 5000 J, m = 4 kg, c = 250 J/kg.°c
Substitute into equation 2
t₁ = 22-[5000/(4×250)
t₁ = 22-(5000/1000)
t₁ = 22-5
t₁ = 17 °C
Explanation:
It is given that,
Speed, v₁ = 7.7 m/s
We need to find the velocity after it has risen 1 meter above the lowest point. Let it is given by v₂. Using the conservation of energy as :




So, the velocity after it has risen 1 meter above the lowest point is 6.26 m/s. Hence, this is the required solution.
Answer: 31.33 degrees
Explanation:
The diffraction angles
when we have a slit divided into
parts are obtained by the following equation:
(1)
Where:
is the width of the slit
is the wavelength of the light
is an integer different from zero.
Now, the first-order diffraction angle is given when
, hence equation (1) becomes:
(2)
Now we have to find the value of
:
(3)
We know:

In addition we are told the diffraction grating has 5000 slits per mm, this means:

Substituting the known values in (3):


<u>Finally:</u>
>>>This is the first-order diffraction angle
Using V= vo +at with Vo = 0 and a= 4m/sec2.
V= 0+ 4x8= 32m/s