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2 years ago

As a projectile moves down its vertical velocity is positive?

1 answer:

7 0

When YOU start working on a problem, if the question doesn't TELL you which direction is positive and which is negative, then YOU get to decide before you start solving the problem.

MOST often, people use positive for UP and negative for DOWN, just because that's the way we naturally think of 'up' and 'down'. But there's no law. If you want to, you can just as well call up 'negative' and call down 'positive'.

You might be interested in

If wavelength and speed of a wave are 4 m and 332 m/s respectively, calculate its frequency

Furkat [3]

Explanation:

Given 

wavelength = 4 m

speed = 332 m/s

frequency = ?

We know we have the formula 

wavelength = speed / frequency 

4 = 332 / frequency 

frequency = 332/4

Therefore frequency is 83 Hertz .

4 0

2 years ago

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

The x vector component of a displacement vector has a magnitude of 146 m and points along the negative x axis. The y vector comp

larisa86 [58]

Answer:

a) the magnitude of r is 184.62

b) the direction is 37.74° south of the negative x-axis

Explanation:

Given the data in the question;

as illustrated in the image blow;

To find the the magnitude of r, we will use the Pythagoras theorem

r² = y² + x²

r = √( y² + x²)

we substitute

r = √((-113)² + (-146)²)

r = √(12769 + 21316 )

r = √(34085 )

r = 184.62

Therefore, the magnitude of r is 184.62

To find its direction, we need to find ∅

from SOH CAH TOA

tan = opposite / adjacent

tan∅ = -113 / -146

tan∅ = 0.77397

∅ = tan⁻¹( 0.77397 )

∅ = 37.74°

Therefore, the direction is 37.74° south of the negative x-axis

7 0

2 years ago

A 1.15 kg book is at rest on the table. What is the magnitude of the normal force that the table is exerting on the book?

Agata [3.3K]

Answer:

11.27N

Explanation:

Given parameters:

Mass of the book = 1.15kg

Unknown:

Magnitude of the normal force = ?

Solution:

The normal force is the vertical force exerted by a body on an object.

It can be described as the weight of an object.

Normal force = mass x acceleration due to gravity

Normal force = 1.15 x 9.8 = 11.27N

5 0

2 years ago

You're driving a vehicle of mass 1350 kg and you need to make a turn on a flat road. The radius of curvature of the turn is 71 m

sergey [27]

Answer:

$v=12.65\ m.s^{-1}$

Explanation:

Given:

mass of vehicle, $m=1350\ kg$

radius of curvature, $r=71\ m$

coefficient of friction, $\mu=0.23$

During the turn to prevent the skidding of the vehicle its centripetal force must be equal to the opposite balancing frictional force:

$m.\frac{v^2}{r} =\mu.N$

where:

$\mu=$ coefficient of friction

$N=$ normal reaction force due to weight of the car

$v=$ velocity of the car

$1350\times \frac{v^2}{71} =0.23\times (1350\times 9.8)$

$v=12.65\ m.s^{-1}$ is the maximum velocity at which the vehicle can turn without skidding.

5 0

3 years ago