Wht is the mass number of 19k/39

Two teams of nine members each engage in a tug of war. Each of the first team’s members has an average mass of 68 kg and exerts

olga2289 [7]

Answer:

a) a = - 0.106 m/s^2 (←)

b) T = 12215.1064 N

Explanation:

If

F₁ = 9*1350 N = 12150 N (→)

F₂ = 9*1365 N = 12285 N (←)

∑Fx = M*a = (M₁ +M₂)*a (→)

F₁ - F₂ = (M₁ +M₂)*a

→ a = (F₁ - F₂) / (M₁ +M₂ ) = (12150-12285)N/(9*68+9*73)Kg

→ a = - 0.106 m/s^2 (←)

(b) What is the tension in the section of rope between the teams?

If we apply ∑Fx = M*a for the team 1

F₁ - T = - M₁*a ⇒ T = F₁ + M₁*a

⇒ T = 12150 N + (9 * 68 Kg)*(0.106 m/s^2)

⇒ T = 12215.1064 N

If we choose the team 2 we get

- F₂ + T = - M₂*a ⇒ T = F₂ - M₂*a

⇒ T = 12285 N - (9 * 73 Kg)*(0.106 m/s^2)

⇒ T = 12215.1064 N

4 0

3 years ago

An electric motor rotating a workshop grinding wheel at 1.00 × 10² rev/min is switched off. Assume the wheel has a constant nega

ololo11 [35]

An electric engine turning a workshop sanding rotation at 1.00 × 10² rev/min is switched off. Take the wheel includes a regular negative angular acceleration of volume 2.00 rad/s². 5.25 moments long it takes the grinding rotation to control.

<h3>What is negative angular acceleration?</h3>

A particle that has a negative angular velocity rotates counterclockwise.
Negative angular acceleration () is a "push" that is hence counterclockwise.
The body will speed up or slow down depending on whether and have the same sign (and eventually go in reverse).
For instance, when an object rotating counterclockwise slows down, acceleration would be negative.
If a rotating body's angular speed is seen to grow in a clockwise direction and decrease in a counterclockwise direction, it is given a negative sign.
It is known that a change in the linear acceleration correlates to a change in the linear velocity.

Let t be the time taken to stop.

ω = 0 rad/s

Use the first equation of motion for rotational motion

ω = ωo + α t

0 = 10.5 - 2 x t

t = 5.25 second

To learn more about angular acceleration, refer to:

brainly.com/question/21278452

#SPJ4

7 0

2 years ago

Which waves are longitudinal waves? Check all that apply.

alekssr [168]

Sound waves.
Eathquake P-Waves

7 0

3 years ago

Read 2 more answers

Two small nonconducting spheres have a total charge of 90.0 C.

valentina_108 [34]

Answer: (a) Smaller charge is $2.7 \times 10^{-5} C$ and larger charge is $11.7 \times 10^{-5} C$ .

(b) Smaller charge is $-11.4 \times 10^{-5}$ and larger charge is $9.1 \times 10^{-5}$ .

Explanation:

(a) When both the spheres have same charge then force is repulsive in nature as like charges tend to repel each other.

Therefore, total charge on the two non-conducting spheres will be calculated as follows.

$Q_{1} + Q_{2} = 90 \mu \times \frac{10^{-6}C}{1 \muC}$

= $9 \times 10^{-5} C$

Therefore, force between the two spheres will be calculated as follows.

F = $k\frac{Q_{1}Q_{2}}{r^{2}}$

12 N = $\frac{(9 \times 10^{9} Nm^{2}/C^{2})Q_{1}Q_{2}}{(0.28 m^{2})}$

$Q_{1}Q_{2} = 0.104 \times 10^{-9} C^{2}$

or, $Q_{1}(9 \times 10^{-5} - Q_{1}) = 0.104 \times 10^{-9} C^{2}$

$9 \times 10^{-5}Q_{1} - Q^{2}_{1} = 0.104 \times 10^{-9} C^{2}$

$Q^{2}_{1} - 9 \times 10^{-5}Q_{1} + 0.104 \times 10^{-9} = 0$

$Q_{1} = 11.7 \times 10^{-5} C, 2.7 \times 10^{-5} C$

This means that smaller charge is $2.7 \times 10^{-5} C$ and larger charge is $11.7 \times 10^{-5} C$ .

(b) When force is attractive in nature then it means both the charges are of opposite sign.

Hence, total charge on the non-conducting sphere is as follows.

$Q_{1} + (-Q_{2}) = 90 \mu \times \frac{10^{-6}C}{1 \muC}$

$Q_{1} - Q_{2} = 9 \times 10^{-5} C$

Now, force between the two spheres is calculated as follows.

F = $k\frac{Q_{1}Q_{2}}{r^{2}}$

12 N = $\frac{(9 \times 10^{9} Nm^{2}/C^{2})Q_{1}Q_{2}}{(0.28 m^{2})}$

$Q_{1}Q_{2} = 0.104 \times 10^{-9} C^{2}$

$Q_{1}(Q_{1} - 9 \times 10^{-5}) = 0.104 \times 10^{-9} C^{2}$

$Q^{2}_{1} - 9 \times 10^{-5}Q_{1} = 0.104 \times 10^{-9} C^{2}$

$Q_{1} = -11.4 \times 10^{-5}, 9.1 \times 10^{-5}$

Hence, smaller charge is $-11.4 \times 10^{-5}$ and larger charge is $9.1 \times 10^{-5}$ .

8 0

3 years ago

A 51-kg packing crate is pulled with constant speed across a rough floor with a rope that is at an angle of 37.0Â° above the hor

fiasKO [112]

work is done by the pulling force which is same as the tension force in the rope. the net work done is zero for the crate since crate moves at constant velocity. but there is work done by the tension force which is equal in magnitude to the work done by the frictional force.

T = tension force in the rope = 115 N

d = displacement of the crate = 7.0 m

θ = angle between the direction of tension force and displacement = 37 deg

work done on the crate is given as

W = F d Cosθ

inserting the values given above

W = (115) (7.0) Cos37

W = 643 J

3 0

3 years ago