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Monica [59]
3 years ago
9

A bus travels 6 km east and then 8 km south. The magnitude of the bus’s resultant displacement is ___km.

Physics
1 answer:
Crank3 years ago
3 0

Answer:

10 km

Explanation:

You can draw your problem out. You can use the pathagoran theorem to figure it out. So 6^{2} + 8^{2} and then square root the answer which will be \sqrt{100} which equals to 10.

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The horizontal beam in Fig. E11.14 weighs 190 N, and its center of gravity is at its center. Find (a) the tension in the cable a
grandymaker [24]

Answer:

(a). The tension in the cable is 658.33 N.

(b). The horizontal components of the force exerted on the beam at the wall is 526.66 N.

(c). The vertical components of the force exerted on the beam at the wall is 95.002 N.

Explanation:

Given that,

Weight of beam= 190 N

Here, The center of gravity is at its center

According to figure,

The angle is

\sin\theta=\dfrac{3}{5}

The horizontal component is

T_{x}=T\cos\theta

The vertical component is

T_{y}=T\sin\theta

(a). We need to calculate the tension in the cable

Using formula of net torque acting on the pivot

\sum\tau=F_{b}\times r+F_{w}\times r'-T\sin \theta\times r'

Put the value into the formula

0=190\times2+300\times 4-T\sin\theta\times 4

T\sin\theta\times 4=380+1200

T=\dfrac{1580\times5}{3\times 4}

T=658.33\ N

(b). We need to calculate the horizontal components of the force exerted on the beam at the wall

Using formula of horizontal component

F_{x}=T\cos\theta

Put the value into the formula

F_{x}=658.33\times\dfrac{4}{5}

F_{x}=526.66\ N

(c). We need to calculate the vertical components of the force exerted on the beam at the wall

Using formula of vertical component

F_{y}=F_{b}+F_{w}-T\sin\theta

Put the value into the formula

F_{y}=190+300-658.33\times\dfrac{3}{5}

F_{y}=95.002\ N

Hence, (a). The tension in the cable is 658.33 N.

(b). The horizontal components of the force exerted on the beam at the wall is 526.66 N.

(c). The vertical components of the force exerted on the beam at the wall is 95.002 N.

3 0
3 years ago
particle of mass 59 g and charge 51 µC is released from rest when it is 32 cm from a second particle of charge −14 µC. Determine
AleksAgata [21]

Answer:

The initial acceleration of the 59g particle is 1062.7\frac{m}{s^{2}}

Explanation:

Newton's second laws relates acceleration (a), net force(F) and mass (m) in the next way:

F=ma (1)

We already know the mass of the particle so we should find the electric force on it to use on (1), the magnitude of the electric force between two charged objects by Columb's law is:

F=k\frac{\mid q_{1}q_{2}\mid}{r^{2}}

with q1 and q2 the charge of the particles, r the distance between them and k the constant k=9.0\times10^{9}\,\frac{Nm^{2}}{C^{2}}. So:

F=(9.0\times10^{9})\frac{\mid (51\times10^{-6})(-14\times10^{-6})\mid}{0.32^{2}}

F=62.7 N

Using that value on (1) and solving for a

a=\frac{F}{m}=\frac{62.7}{0.059}=1062.7\frac{m}{s^{2}}

4 0
3 years ago
In a doorknob, the knob is connected to a shaft. When the knob turns, the shaft turns,which moves the door latch. The radius of
HACTEHA [7]

Answer:

`7

Explanation:

i know

4 0
2 years ago
PLEASE HELP!! ITS DUE AT 11:59!!
DENIUS [597]

Answer:

14.1 po

sana makatulong

3 0
3 years ago
3. A sprinter leaves the starting blocks with an acceleration of 4.5 m/s2. What is the
UkoKoshka [18]

Hi there! :)

\large\boxed{v_{f} = 18 m/s}

Use the following kinematic equation to solve for the final velocity:

v_{f} = v_{i} + at

In this instance, the runner started from rest, so the initial velocity is 0 m/s. We can rewrite the equation as:

v_{f} = at

Plug in the given acceleration and time:

v_{f} = 4.5 * 4 = 18 m/s

5 0
3 years ago
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