Answer:
Specific heat of brass is 0.40 J g⁻¹ °C⁻¹ .
Explanation:
Given :
Mass of brass, m₁ = 440 g
Temperature of brass, T₁ = 97° C
Mass of water, m₂ = 350 g
Temperature of water, T₂ = 23° C
Specific heat of water, C₂ = 4.18 J g⁻¹ °C⁻¹
Equilibrium temperature, T = 31° C
Let C₁ be the specific heat of brass.
Heat loss by brass = Heat gain by water
m₁ x C₁ x ( T₁ -T ) = m₂ x C₂ x ( T - T₁ )
Substitute the suitable values in above equation.
440 x C₁ x (97 - 31) = 350 x 4.18 x (31 - 23)
C₁ = 
C₁ = 0.40 J g⁻¹ °C⁻¹
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Answer:

Explanation:
Given data:
PERIOD OF MOTION IS T = 25.5 days
orbital speeds = 220 km/s
we know that
acceleration due to centripetal force is
Gravitational force
we know that

solving for





we know that
f =ma

solving for m



Answer:
Bone
Explanation:
Diagnostic radiology include the use of non-invasive imaging scans to diagnose a patient.
The voltages used in diagnostic tubes range from roughly 20 kV to 150 kV and thus the highest energies of the X-ray photons range from roughly 20 keV to 150 keV.
The tests and equipment used sometimes involves low doses of radiation to create highly detailed images of an area.