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3 years ago

At one moment during a walk around the block, there are four forces exerted upon Fido - a 10.0 kg dog. The forces are:

Physics

1 answer:

Ann [662]3 years ago

7 0

Explanation:

Given that,

Fapp = 67.0 N at 30.0° above the horizontal (rightward and upward)

Fnorm = 64.5 N, up

Ffrict = 27.6 N, left

Fgrav = 98 N, down

(a) Resolution of the applied force:

Horizontal component,

$F_x=F\cos\theta\\\\=67\times \cos(30)\\\\=58.02\ N$

Vertical component,

$F_y=F\sin\theta\\\\=67\times \sin(30)\\\\=33.5\ N$

(b) Net horizontal force :

$F_x=58.02+(-27.6),\ \text{frictional force act in opposite direction of motion}\\\\F_x=30.42\ N$

It is positive, it will act in right side

Net vertical force :

$F_y=33.5+64.5+(-98),\ \text{gravitational force acts in downward direction}\\\\F_y=0$

Hence, it is clear that the net force is in horizontal direction i.e. 30.42 N due right side.

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A projectile is shot from the edge of a cliff above the ground level with initial velocity of at an angle with the horizontal. (

Xelga [282]

Answer:

t = √2y/g

Explanation:

This is a projectile launch exercise

a) The vertical velocity in the initial instants ( $v_{oy}$ = 0) zero, so let's use the equation

y = $v_{oy}$ t -1/2 g t²

y= - ½ g t²

t = √2y/g

b) Let's use this time and the horizontal displacement equation, because the constant horizontal velocity

x = vox t

x = v₀ₓ √2y/g

c) Speeds before touching the ground

vₓ = vox = constant

$v_{y}$ = $v_{oy}$ - gt

$v_{y}$ = 0 - g √2y/g

$v_{y}$ = - √2gy

tan θ = Vy / vx

θ = tan⁻¹ (vy / vx)

θ = tan⁻¹ (√2gy / vox)

d) The projectile is higher than the cliff because it is a horizontal launch

6 0

3 years ago

Which is an example of an exothermic process?

Scorpion4ik [409]

Answer:

an example of an exthermic process is combustion

Explanation:

combustion is like lighting a candle

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3 years ago

A 560-g squirrel with a surface area of 930cm2 falls from a 5.0-m tree to the ground. Estimate its terminal velocity. (Use a dra

bija089 [108]

Explanation:

Terminal velocity is given by:

$v_t=\sqrt{\frac{2mg}{\rho C_dA}}$

Here, m is the mass of the falling object, g is the gravitational acceleration, $C_d$ is the drag coefficient, $\rho$ is the fluid density through which the object is falling, and A is the projected area of the object. in this case the projected area is given by:

$A=\frac{A_s}{2}=\frac{930cm^2}{2}=465cm^2\\465cm^2*\frac{1m^2}{10^4cm^2}=0.0465m^2\\560g*\frac{1kg}{10^3g}=0.56kg$

Recall that drag coefficient for a horizontal skydiver is equal to 1 and air density is $1.28\frac{kg}{m^3}$ .

$v_t=\sqrt{\frac{2(0.56kg)(9.8\frac{m}{s^2})}{(1.28\frac{kg}{m^3}(1)(0.0465m^2)}}\\v_t=13.58\frac{m}{s}$

Without drag contribution the motion of the person is an uniformly accelerated motion, thus:

$v_f^2=v_o^2+2gh\\v_f=\sqrt{2gh}\\v_f=\sqrt{2(9.8\frac{m}{s^2})(5m)}\\v_f=9.9\frac{m}{s}$

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3 years ago

Why would genetic material need to be able to reproduce

Lynna [10]

So that the next generation could inherit the previous adaptations and instinct and be similar to the parents. if that didn't happen, a cat might give birth to a giraffe

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4 years ago

The net force acting on the ball below is ___ N

yKpoI14uk [10]

The net force applied to the object equals the mass of the object multiplied by the amount of its acceleration." The net force acting on the soccer ball is equal to the mass of the soccer ball multiplied by its change in velocity each second (its acceleration).

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3 years ago