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anastassius [24]
3 years ago
15

(Double points) A worker applied 27 newtons to a lever with a length of 4 meters that rotated around a hinge. What was the torqu

e?
Physics
2 answers:
arlik [135]3 years ago
7 0

Answer: Torque T = 108 Nm

Explanation:

Torque or turning effect is the rotational equivalent of linear force, i.e moment of force about a point.

Given that the force is applied perpendicular to the edge (angle 90°)

Torque can be represented by T;

T = FrSin∅ = Fr ( sin90 = 1)

T = force × distance

T = F × r

F = 27N

r = 4m

T = 27N × 4m

T = 108 Nm

faltersainse [42]3 years ago
4 0

Answer:

The torque was 108 N m

Explanation:

Torque is the cross product between the vector position where the force is applied with respect the axis of rotation and the force applied:

\overrightarrow{\tau}=\overrightarrow{r}\times\overrightarrow{F}

The magnitude is:

\tau =rF\sin \theta

If the force is applied on an extreme of the lever, the hinge is in the other extreme and the applied force is perpendicular to the lever:

r=4m, F=27N, \theta=90 Celcius

So:

\tau=(4m)(27N)\sin (90)=(108Nm)(1)

\tau=108 Nm

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Answer:

The minimum time, t = 21.11 m

Explanation:

Given,

The maximum distance between the Earth and Mars, d = 780,000,000 km

The maximum distance between the Earth and Mars, d = 780,000,000 km

The speed of the light, c = 300,000 km/s

The distance and speed of the object is related using the formula,

                           v = d / t

                            t = d / v

The minimum time it takes for data transmission is when the distance between the Mars and the Earth is at minimum.

Therefore,

                            t = 380,000,000 km / 300,000 km/s

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Hence, the minimum time it takes for data transmitted by the Mars Surveyor to reach earth, t = 21.11 minutes

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3. Discuss the main purpose of Experiment" Explain the greatest advantage of experiment in psychology?
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3 years ago
. Consider the equation =0+0+02/2+03/6+04/24+5/120, where s is a length and t is a time. What are the dimensions and SI units of
Olegator [25]

Answer:

See Explanation

Explanation:

Given

s=s_0+v_0t+\frac{a_0t^2}{2}+ \frac{j_0t^3}{6}+\frac{S_0t^4}{24}+\frac{ct^5}{120}

Solving (a): Units and dimension of s_0

From the question, we understand that:

s \to L --- length

t \to T --- time

Remove the other terms of the equation, we have:

s=s_0

Rewrite as:

s_0=s

This implies that s_0 has the same unit and dimension as s

Hence:

s_0 \to L --- dimension

s_o \to Length (meters, kilometers, etc.)

Solving (b): Units and dimension of v_0

Remove the other terms of the equation, we have:

s=v_0t

Rewrite as:

v_0t = s

Make v_0 the subject

v_0 = \frac{s}{t}

Replace s and t with their units

v_0 = \frac{L}{T}

v_0 = LT^{-1}

Hence:

v_0 \to LT^{-1} --- dimension

v_0 \to m/s --- unit

Solving (c): Units and dimension of a_0

Remove the other terms of the equation, we have:

s=\frac{a_0t^2}{2}

Rewrite as:

\frac{a_0t^2}{2} = s_0

Make a_0 the subject

a_0 = \frac{2s_0}{t^2}

Replace s and t with their units [ignore all constants]

a_0 = \frac{L}{T^2}\\

a_0 = LT^{-2

Hence:

a_0 = LT^{-2 --- dimension

a_0 \to m/s^2 --- acceleration

Solving (d): Units and dimension of j_0

Remove the other terms of the equation, we have:

s=\frac{j_0t^3}{6}

Rewrite as:

\frac{j_0t^3}{6} = s

Make j_0 the subject

j_0 = \frac{6s}{t^3}

Replace s and t with their units [Ignore all constants]

j_0 = \frac{L}{T^3}

j_0 = LT^{-3}

Hence:

j_0 = LT^{-3} --- dimension

j_0 \to m/s^3 --- unit

Solving (e): Units and dimension of s_0

Remove the other terms of the equation, we have:

s=\frac{S_0t^4}{24}

Rewrite as:

\frac{S_0t^4}{24} = s

Make S_0 the subject

S_0 = \frac{24s}{t^4}

Replace s and t with their units [ignore all constants]

S_0 = \frac{L}{T^4}

S_0 = LT^{-4

Hence:

S_0 = LT^{-4 --- dimension

S_0 \to m/s^4 --- unit

Solving (e): Units and dimension of c

Ignore other terms of the equation, we have:

s=\frac{ct^5}{120}

Rewrite as:

\frac{ct^5}{120} = s

Make c the subject

c = \frac{120s}{t^5}

Replace s and t with their units [Ignore all constants]

c = \frac{L}{T^5}

c = LT^{-5}

Hence:

c \to LT^{-5} --- dimension

c \to m/s^5 --- units

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3 years ago
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REY [17]

The correct answer is

C. The battery

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8 0
3 years ago
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