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3 years ago

1. A group of students were trying to find the greatest

Physics

1 answer:

deff fn [24]3 years ago

8 0

Answer:

D th

Explanation:

D B. The force applied to the ball is a balanced force.

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ASAP pls answer right if can’t see picture don’t answer

Anna35 [415]

Answer:

Not sure but

F = m* a

32= 5 * a

a= 6.4 m/s^2

6 0

3 years ago

video for A door 1 m wide, of mass 15 kg, is hinged at one side so that it can rotate without friction about a vertical axis. It

natka813 [3]

Answer:

$\omega_f = 0.4\ rad/s$

Explanation:

given,

width of door dimension = 1 m

mass of door = 15 Kg

mass of bullet = 10 g = 0.001 Kg

speed of bullet = 400 m/s

$I_{total} =I_{door} + I_{bullet}$

$I_{total} =\dfrac{1}{3}MW^2 + m(\dfrac{W}{2})^2$

a) from conservation of angular momentum

$L_i = L_f$

$mv\dfrac{W}{2} = I_{total}\omega_f$

$mv\dfrac{W}{2}= (\dfrac{1}{3}MW^2 + m(\dfrac{W}{2})^2)\omega_f$

$\omega_f= \dfrac{mv\dfrac{W}{2}}{\dfrac{1}{3}MW^2 + m(\dfrac{W}{2})^2}$

$\omega_f= \dfrac{\dfrac{mv}{2}}{\dfrac{MW }{3}+(\dfrac{mW}{4})}$

$\omega_f= \dfrac{\dfrac{0.01\times 400}{2}}{\dfrac{15\times 1 }{3}+(\dfrac{0.01\times 1}{4})}$

$\omega_f = 0.4\ rad/s$

8 0

3 years ago

A 81 kg person sits on a 3.8 kg chair. Each leg of the chair makes contact with the floor in a circle that is 1.2 cm in diameter

tankabanditka [31]

Answer:

1.9 MPa

Explanation:

Mass of person = 81 kg

Mass of chair = 3.8 kg

Diameter of contact point = 1.2 cm = D

Radius of contact point = 1.2/2 = 0.6 cm

Total mass of chair and person = 81 + 3.8 = 84.8 kg = M

Acceleration due to gravity = 9.81 m/s²

Force acting on the floor

F = Mg

⇒F = 84.8×9.81

⇒F = 831.888 N

Area of the contact point

A = πR²

⇒A = π0.006²

⇒A = π0.000036 m²

Area of the four points is

4A = 0.000144π m²

Pressure

$p=\frac{F}{A}\\\Rightarrow p=\frac{831.888}{0.000144\pi}\\\Rightarrow p=1838876.21\ Pa=1.83887621\times 10^6\ Pa=1.9 MPa$

Pressure exerted on the floor by each leg of the chair is 1.9 MPa

5 0

3 years ago

A 2000 kg truck traveling north at 34 km/h turns east and accelerates to 58 km/h. (a) What is the change in the truck's kinetic

barxatty [35]

Explanation:

It is given that,

Mass of the truck, m = 2000 kg

Initial velocity of the truck, u = 34 km/h = 9.44 m/s

Final velocity of the truck, v = 58 km/h = 16.11 m/s

(a) Change in truck's kinetic energy, $\Delta E=\dfrac{1}{2}m(v^2-u^2)$

$\Delta E=\dfrac{1}{2}\times 2000\ kg\times (16.11^2-9.44^2)$

$\Delta E=170418.5\ J$

$\Delta E=1.7\times 10^5\ J$

(b) Change in momentum of the truck, $\Delta p=m(v-u)$

$\Delta p=2000\ kg\times (16.11-9.44)$

$\Delta p=13340\ kg-m/s$

Hence, this is the required solution.

6 0

3 years ago

Calculate the wavelength of an electron (m = 9.11 × 10-28 g) moving at 3.66 × 106 m/s.

valina [46]

The answer is 1.99 × 10⁻¹⁰ m.

To calculate this we will use De Broglie wavelength formula:
λ = h/(m*v)
λ - the wavelength
h - Plank's constant: h = 6.626 × 10⁻³⁴ Js
v - speed
m - mass

It is given:
λ = ?
m = 9.11 × 10⁻²⁸ g
v = 3.66 × 10⁶ m/s

After replacing in the formula:
λ = h/(m*v) = 6.626 × 10⁻³⁴ /(9.11 × 10⁻²⁸ * 3.66 × 10⁶) = 1.99 × 10⁻¹⁰ m

6 0

3 years ago

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