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kaheart [24]
3 years ago
7

What is the weight of a 1-kilogram brick resting on a table?

Physics
1 answer:
MakcuM [25]3 years ago
5 0

Answer:

The weight if the block is 10Newtons

Explanation:

The weight of any object is quantity of matter the object contains and it is always acting downwards on such body. This shows that the object is under the influence of gravity.

The weight of an object is calculated as mass of the object × its acceleration due to gravity

W = mg

Give the mass of the brick to be 1kg

g is the acceleration due to gravity = 10m/s²

Weight of the object = 1 × 10

= 10kgm/s² or 10Newtons

You might be interested in
A solid cylinder of mass M = 45 kg, radius R = 0.44 m and uniform density is pivoted on a frictionless axle coaxial with its sym
user100 [1]

Answer:

w_f = 1.0345 rad/s

Explanation:

Given:

- The mass of the solid cylinder M = 45 kg

- Radius of the cylinder R = 0.44 m

- The mass of the particle m = 3.6 kg

- The initial speed of cylinder w_i = 0 rad/s

- The initial speed of particle V_pi = 3.3 m/s

- Mass moment of inertia of cylinder I_c = 0.5*M*R^2

- Mass moment of inertia of a particle around an axis I_p = mR^2

Find:

- What is the magnitude of its angular velocity after the collision?

Solution:

- Consider the mass and the cylinder as a system. We will apply the conservation of angular momentum on the system.

                                     L_i = L_f

- Initially, the particle is at edge at a distance R from center of cylinder axis with a velocity V_pi = 3.3 m/s contributing to the initial angular momentum of the system by:

                                    L_(p,i) = m*V_pi*R

                                    L_(p,i) = 3.6*3.3*0.44

                                    L_(p,i) = 5.2272 kgm^2 /s

- While the cylinder was initially stationary w_i = 0:

                                    L_(c,i) = I*w_i

                                    L_(c,i) = 0.5*M*R^2*0

                                    L_(c,i) = 0 kgm^2 /s

The initial momentum of the system is L_i:

                                    L_i = L_(p,i) + L_(c,i)

                                    L_i = 5.2272 + 0

                                    L_i = 5.2272 kg-m^2/s

- After, the particle attaches itself to the cylinder, the mass and its distribution around the axis has been disturbed - requires an equivalent Inertia for the entire one body I_equivalent. The final angular momentum of the particle is as follows:

                                   L_(p,f) = I_p*w_f

- Similarly, for the cylinder:

                                   L_(c,f) = I_c*w_f

- Note, the final angular velocity w_f are same for both particle and cylinder. Every particle on a singular incompressible (rigid) body rotates at the same angular velocity around a fixed axis.

                                  L_f = L_(p,f) + L_(c,f)

                                  L_f = I_p*w_f + I_c*w_f

                                  L_f = w_f*(I_p + I_c)

-Where, I_p + I_c is the new inertia for the entire body = I_equivalent that we discussed above. This could have been determined by the superposition principle as long as the axis of rotations are same for individual bodies or parallel axis theorem would have been applied for dissimilar axes.

                                  L_i = L_f

                                  5.2272 = w_f*(I_p + I_c)

                                  w_f =  5.2272/ R^2*(m + 0.5M)

Plug in values:

                                  w_f =  5.2272/ 0.44^2*(3.6 + 0.5*45)

                                  w_f =  5.2272/ 5.05296

                                  w_f = 1.0345 rad/s

5 0
3 years ago
Sara Lilia pulled out four different CD's to find out which type helped her to finish her
lord [1]
Step by step explanation
7 0
3 years ago
An air bubble has a volume of 1.3 cm3 when it is released by a submarine 160 m below the surface of a freshwater lake. What is t
murzikaleks [220]

Answer:

V2 = 21.44cm^3

Explanation:

Given that: the initial volume of the bubble = 1.3 cm^3

Depth = h = 160m

Where P2 is the atmospheric pressure = Patm

P1 is the pressure at depth 'h'

Density of water = ρ = 10^3kg/m^3

Patm = 1.013×10^5 Pa.

Patm = 101300Pa

g = 9.81m/s^2

P1 = P2+ρgh

P1 = Patm +ρgh

P1 = 1.013×10^5+10^3×9.81×160.

P1 = 101300+1569600

P1 = 1670900 Pa

For an ideal gas law

PV =nRT

P1V1/P2V2 = 1

V2 = ( P1/P2)V1

V2 = (P1/Patm)V1

V2 = ( 1670900 /101300 Pa) × 1.3

V2 = 1670900/101300

V2 = 16.494×1.3

V2 = 21.44cm^3

4 0
3 years ago
Consider the video you just watched. Suppose we replace the original launcher with one that fires the ball upward at twice the s
sveta [45]

Answer:

b

Explanation:

Given:

- The ball is fired at a upward initial speed v_yi = 2*v

- The ball in first experiment was fired at upward initial speed v_yi = v

- The ball in first experiment was as at position behind cart = x_1

Find:

How far behind the cart will the ball land, compared to the distance in the original experiment?

Solution:

- Assuming the ball fired follows a projectile path. We will calculate the time it takes for the ball to reach maximum height y. Using first equation of motion:

                                      v_yf = v_yi + a*t

Where, a = -9.81 m/s^2 acceleration due to gravity

            v_y,f = 0 m/s max height for both cases:

For experiment 1 case:

                                     0 = v - 9.81*t_1

                                      t_1 = v / 9.81

For experiment 2 case:

                                     0 = 2*v - 9.81*t_2

                                      t_2 = 2*v / 9.81

The total time for the journey is twice that of t for both cases:

For experiment 1 case:

                                     T_1 = 2*t_1

                                     T_1 = 2*v / 9.81

For experiment 2 case:

                                     T_2 = 2*t_2

                                     T_2 = 4*v / 9.81

- Now use 2nd equation of motion in horizontal direction for both cases:

                                     x = v_xi*T

For experiment 1 case:

                                     x_1 = v_x1*T_1

                                    x_1 = v_x1*2*v / 9.81

For experiment 2 case:

                                     x_2 =  v_x2*T_2

                                    x_2 = v_x2*4*v / 9.81

- Now the x component of the velocity for each case depends on the horizontal speed of the cart just before launching the ball. Using conservation of momentum we see that both v_x2 = v_x1 after launch. Since the masses of both ball and cart remains the same.

- Hence; take ratio of two distances x_1 and x_2:

                        x_2 / x_2 = v_x2*4*v / 9.81 * 9.81 / v_x1*2*v

Simplify:

                        x_1 / x_2 = 2  

- Hence, the amount of distance traveled behind the cart in experiment 2 would be twice that of that in experiment 1.      

                                   

3 0
3 years ago
HELP PLEASE GIVING 30 POINTS AND BRAINLIEST DUE VERY SOON!!
olga55 [171]

Answer:

3 is your anwser

Explanation:

5 0
3 years ago
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