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3 years ago

14

g A rod is 2.0 m long and lies along the x-axis, with one end at the origin. A force of 25 N is applied at the point x = 1.2 m,

and is directed 30° above the x-axis. What is the torque on the rod?

1 answer:

Nataly [62]3 years ago

6 0

Answer:

τ = 15N.m

Explanation:

Torque on the rod is given by:

$\tau=F*sin(30\°)*d$

$\tau=25*0.5*1.2$

$\tau=15N.m$

If you were told that gravity contributes as well, and you were given the mass of the rod, you would have to subtract the torque of the weight to this value.

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A car travels in a straight line for 5 h at a constant speed of 72 km/h. What is it’s acceleration

Luda [366]

Answer:

$a \approx \: 0.001 \: m {s}^{ - 2}$

Explanation:

Given:

$initial \: velocity \: (u) = 0 \\ \\ Final \: Velocity \: (v) = 72 km /h \\ \\ Time \: (t) = 5 \: hours \\ \\ Acceleration \: (a) =? \\ \\ \because \: a = \frac{v - u}{t} \\ \\ \therefore \: a = \frac{72 - 0}{5} \\ \\ a = \frac{72}{5} \\ \\ a = 14.5 \: km {h}^{ - 2} \\ \\ a = \frac{14.5 \times 1000}{3600\times 3600} \: m {s}^{ - 2} \\ \\ a = 0.00111882716 \\ \\ a \approx \: 0.001 \: m {s}^{ - 2}$

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3 years ago

3. Why is static electricity not useful as a power

shtirl [24]

The answer is B, because all energy is released at once in static electricity.

There’s a quizlet that mentions these questions, if you are having trouble. I’d suggest to give them a look.

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2 years ago

2. Our solar system is made up of the Sun, 8 planets, and other bodies such as

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It’s 25 because I already took the testing

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2 years ago

At the instant the traffic light turns green, an automobile that has been waiting at an intersection starts ahead with a constan

oksian1 [2.3K]

Answer:

(A) Distance taken by automobile to overtake truck is 175.23 m

(B) Final speed of the automobile is 29.6 m/s.

Explanation:

According to the problem, the automobile is at rest, so its initial speed is zero and the automobile is moving with constant acceleration.

The general equation of motion is:

d = ut + 0.5at² .....(1)

Here d is distance, a is acceleration, t is time and u is initial speed.

The distance (d₁) travel by the automobile after green light in time t is given by the relation :

d₁ = u₁t + 0.5at²

We know that its initial velocity is zero i.e. u = 0 m/s.

d₁ = 0.5at² ......(2)

The truck is moving with constant speed, so its acceleration is zero. Therefore, the distance covered by the truck using equation (1) is;

d₂ = u₂t .......(3)

Since, the distance travelled by automobile and truck are same. Hence,

d₁ = d₂

0.5at² = u₂t

t = 2u₂/a

Substitute 14.8 m/s for u₂ and 2.50 m/s² for a in the above equation.

t = $\frac{2\times14.8}{2.50}$

t = 11.84 s

(A) Distance travelled by automobile, d = u₂t =14.8 x 11.84 = 175.23 m

(B) Final speed of the automobile is given by the relation:

v = u + at

v = 0 + 2.50 x 11.84

v = 29.6 m/s

4 0

3 years ago

Two ice skaters collide on the ice. A 39.6-kg skater moving South at 6.21 m/s collides with a 52.1-kg skater moving East at 4.33

Stells [14]

Answer:

V = 3.6385 m/s

θ = 47.46 degrees

Explanation:

the important data in the question is:

Skater 1:

$M_1$ = 39.6 kg

direction: south (axis y)

$V_{1iy}$ = 6.21 m/s

Skater 2:

$M_2$ = 52.1 kg

direction: east (axis x)

$V_{2ix}$ = 4.33 m/s

Now using the law of the conservation of linear momentum ( $P_i = P_f$ and knowing that the collision is inelastic we can do the next equations:

$M_{1}V_{1ix}+M_2V_{2ix} = V_{sx}(M_1+M_2)$ (eq. 1)

$M_{1}V_{1iy}+M_2V_{2iy} = V_{sy}(M_1+M_2)$ (eq. 2)

Where $V_{sx}$ and $V_{sy}$ is the velocity of the sistem in x and y after the collision.

Note: the conservation of the linear momentum have to be make once by each axis.

Now, in the (eq. 1) the skater 1 don't have velocity in the axis x, so we can replace $V_{1ix}$ by 0 in the equation and get:

$M_2V_{2ix} = V_{sx}(M_1+M_2)$ (eq. 1)

also, in the (eq. 2) the skater 2 don't have velocity in the axis y, so we can replace $V_{2iy}$ by 0 in the equation and get:

$M_{1}V_{1iy} = V_{sy}(M_1+M_2)$ (eq. 2)

Now, we just replace the data in both equations:

$(52.1)(4.33) = V_{sx}(39.6+52.1)$ (eq. 1)

$(39.6)(6.21) = V_{sy}(39.6+52.1)$ (eq. 2)

solving for $V_{sx]$ and $V_{sy}$ we have:

$V_{sx]$ = 2.46 m/s

$V_{sy]$ = 2.681 m/s

using the pythagoras theorem we can find the magnitude of the velocity as:

V = $\sqrt{2.46^2+2.681^2}$

V = 3.6385 m/s

For find the direction we just need to do this;

θ = $tan^{-1}(\frac{2.681}{1.46})$

θ = 47.46 degrees

3 0

3 years ago