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2 years ago

if a cyclist traveled southwest a distance of 12,250 meters in 1.7 hrs what would the velocity of the cyclist be?

Physics

1 answer:

Paladinen [302]2 years ago

7 0

Answer:

Velocity = 7.2 km/hr SW

Explanation:

Velocity is a vector that includes both speed and direction. Speed is calculated by dividing the distance travelled by the time required for that distance.

Speed = Distance/Time

i'll convert 12,250 meters into 12.250 kilometers (km). Not needed, but it's my preference. Time is 1.7 hours:

Speed = (12.250 km)/(1.7 hr) = 7.2 km/hr (to 2 sig figs)

The direction is southwest, or SW

Velocity = 7.2 km/hr SW

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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.

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3 years ago

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sveta [45]

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3 years ago

A(n) 83 kg fisherman jumps from a dock into a 139 kg rowboat at rest on the West side of the dock. If the velocity of the fisher

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