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3 years ago

The diagram shows a handle with three forces, each 100 N, applied to it. The handle is free to

Physics

1 answer:

wolverine [178]3 years ago

6 0

Answer:

с The handle will turn anticlockwise (to the left).

Explanation:

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Which one is the right andwerrrrrrrr

Elden [556K]

2nd one I believe! If I’m wrong sorry

7 0

3 years ago

If the constant acceleration continues for 10.0s, what will be it’s velocity then?

velikii [3]

There's not enough information in the question to find the object's velocity. The best we can do is describe its speed.

The object's speed at the end of the 10.0 seconds =

(its speed at the beginning of the 10.0 seconds)

plus

(10) times (the value of the acceleration) .

5 0

3 years ago

Read 2 more answers

How do I find the tension? its supposed to be in the mid 40s

babunello [35]

Answer:

a = 2.3 m/s²

T = 45 N

Explanation:

Draw a free body diagram for each mass.

For the mass on the incline, there are four forces:

Weight force mg pulling down.

Normal force N perpendicular to the incline.

Friction force Nμ pushing down the incline.

Tension force T pulling up the incline.

For the hanging mass, there are two forces:

Weight force Mg pulling down.

Tension force T pulling up.

Sum of the forces on the hanging mass in the -y direction:

∑F = ma

Mg − T = Ma

T = Mg − Ma

Sum of the forces on the sliding mass in the perpendicular direction:

∑F = ma

N − mg cos θ = 0

N = mg cos θ

Sum of the forces on the sliding mass in the parallel direction:

∑F = ma

T − mg sin θ − Nμ = ma

Substitute:

Mg − Ma − mg sin θ − mgμ cos θ = ma

Mg − mg (sin θ + μ cos θ) = ma + Ma

Mg − mg (sin θ + μ cos θ) = (m + M) a

a = [ Mg − mg (sin θ + μ cos θ) ] / (m + M)

Plug in values:

a = [ 6.0×9.8 − 5.0×9.8 (sin 30° + 0.20 cos 30°) ] / (5.0 + 6.0)

a = 2.3 m/s²

Now find tension:

T = Mg − Ma

T = 6.0×9.8 − 6.0×2.3

T = 45 N

3 0

4 years ago

An RC circuit consists of a resistor with resistance 1.0 kΩ, a 120-V battery, and two capacitors, C1 and C2, with capacitances o

Serhud [2]

Answer:

$Q_t= 8.3 * 10^3 C$

Explanation:

From the question we are told that:

Resistor $R=1000ohms$

Voltage $v=120_V$

Capacitance of c_1 $c_1=20 \mu F$

Capacitance of c_2 $c_2=60 \mu F$

Time $t=0$

Generally the equation for charges is mathematically given by

$For C_1\\Charge\ on\ C_1 = CV = 20*120 = 2400 μC = 2.4 x 10^-3 C\\Charge\ on\ C_1 = 2400 μC = 2.4 x 10^-3 C\\Charge\ on\ C_1 = 2.4 x 10^-3 C\\$

$ForC_2\\Charge on C_2 = 60*120 =7200 μC = 7.2 x 10^-3\\Charge on C_2 = 7.2 x 10^-3$

Generally the equation for voltage across capacitors is mathematically given by

$V_c(t)=V(1-e^{-t/RC})$

$C=C_1+C_2=80 \mu f\\t=2RC=>160000s$

$V_c(t)=120(1-e^{-(160000)/1000*(80)})$

$V_c(t)=103.7598$

Generally the equation for charges is mathematically given by

$Q1(t) = C1Vc(t)\\Q1(t) = 20*103.7598\\Q1(t) = 2075.196\\\\Q2(t) = 60*103.7598\\Q2(t) = 6225.6\\$

Generally the equation for total charges $Q_t$ is mathematically given by

$Q_t=Q1(t)+Q2(t)$

$Q_t= 8.3 * 10^3 C$

5 0

3 years ago

Seawater is full of moving molecules that possess kinetic energy. Could we extract some of this energy

tigry1 [53]

I don’t think soooooo

8 0

3 years ago