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3 years ago

A 2500-kg car is being pushed up a hill at an angle of 35 degrees. Determine the gravitational

Physics

1 answer:

velikii [3]3 years ago

3 0

Answer:

The force that pulls the car down is Wₓ = 14052.6 N and the one that pushes the car up is F = 26552.6 N

Explanation:

For this exercise we will use Newton's second law, let's set a reference system where the x axis is parallel to the plane, in the adjoint we can see the forces in the system.

sin 35 = Wₓ / W

cos 35 = W_y / W

Wₓ = W sin 35

W_y = W cos 35

Wₓ = 2500 9.8 sin 35

Wₓ = 14052.6 N

let's write the equations for each axis

and

Y axis

N-W_y = 0

N = W_y

X axis

F -Wₓ = m a

F = Wₓ + m a = mg sin 35 + m a

F = m (a + g sin 35)

let's calculate

F = 2500 (5 + 9.8 sin 35)

F = 26552.6 N

The force that pulls the car down is Wₓ = 14052.6 N and the one that pushes the car up is F = 26552.6 N

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object P1 and P2 are in a straight line with the normal to a plane mirror.If P1 and P2 are 18m and 21m away from the mirror. Cal

VARVARA [1.3K]

The distance between object P1 and its image formed is determined as 36 m.

<h3>Distance of the image</h3>

The distance of the image formed by object P1 is calculated as follows;

In a plane mirror; object distance = image distance

image distance of P1 = 18 m

distance between object and image = 18m + 18 m = 36 m

Thus, the distance between object P1 and its image formed is determined as 36 m.

Learn more about plane mirrors here: brainly.com/question/1126858

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2 years ago

A 3.0 kg rifle is held firmly by a 50.0 kg woman, initially standing still. A 0.06 kg bullet leaves the rifle muzzle with a velo

ohaa [14]

Answer:

48kg

Explanation:

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3 years ago

You build a grandfather clock, whose timing is based on a pendulum. You measure its period to be 2s on Earth. You then travel wi

Elenna [48]

Answer:

$\frac{g_{2}}{g_{1}} = \frac{1}{4}$

Explanation:

The period of the simple pendulum is:

$T = 2\pi\cdot \sqrt{\frac{l}{g} }$

Where:

$l$ - Cord length, in m.

$g$ - Gravity constant, in $\frac{m}{s^{2}}$ .

Given that the same pendulum is test on each planet, the following relation is formed:

$T_{1}^{2}\cdot g_{1} = T_{2}^{2}\cdot g_{2}$

The ratio of the gravitational constant on planet CornTeen to the gravitational constant on planet Earth is:

$\frac{g_{2}}{g_{1}} = \left(\frac{T_{1}}{T_{2}} \right)^{2}$

$\frac{g_{2}}{g_{1}} = \left(\frac{2\,s}{4\,s} \right)^{2}$

$\frac{g_{2}}{g_{1}} = \frac{1}{4}$

5 0

4 years ago

if vector u has lenght 70 and direction 40 degrees, and vector v has length 85 and direction 335 degrees what is the length and

Anastaziya [24]

Answer:

Magnitude of resultant = 131.15

Direction of resultant = 3.97°

Explanation:

||u|| = 70

θ = 40°

$\vec{u}_x=||u||cos\theta \\\Rightarrow \vec{u}_x=70cos40=53.62$

$\vec{u}_y=||u||sin\theta \\\Rightarrow \vec{u}_y=70sin40=44.99$

||v|| = 85

θ = 335°

$\vec{v}_x=||v||cos\theta \\\Rightarrow \vec{v}_x=85cos335=77.03$

$\vec{v}_y=||v||sin\theta \\\Rightarrow \vec{v}_y=85sin335=-35.92$

Resultant

$R=\sqrt{R_x^2+R_y^2}\\\Rightarrow R=\sqrt{(\vec{u}_x+\vec{v}_x)^2+(\vec{u}_y+\vec{v}_y)^2}\\\Rightarrow R =\sqrt{(70cos40+85cos335)^2+(70sin40+85sin335)^2}\\\Rightarrow R =131.15$

$\theta=tan^{-1}\frac{R_y}{R_x}\\\Rightarrow \theta=tan^{-1}\frac{70sin40+85sin335}{70cos40+85cos335}\\\Rightarrow \theta=tan^{-1}0.069=3.97^{\circ}$

Magnitude of resultant = 131.15

Direction of resultant = 3.97°

4 0

3 years ago

A bicyclist travels 4.5 km west, then travels 6.7 km at an angle 27.0 degrees South of West. What is the magnitude of the bicycl

Dimas [21]

Answer:

Explanation:

Magnitude of the total displacement is expressed using the equation

d = √dx²+dy²

dx is the horizontal component of the displacement

dy is the vertical component of the displacement

dy = -6.7sin27°

dy = -6.7(0.4539)

dy = -3.042

For the horizontal component of the displacement

dx = -4.5 - 6.7cos27

dx = -4.5 -5.9697

dx = -10.4697

Get the magnitude of the bicyclist's total displacement

Recall that: d = √dx²+dy²

d = √(-3.042)²+(-10.4697)²

d = √9.2538+109.6146

d = √118.8684

<em>Hence the magnitude of the bicyclist's total displacement is 10.90km</em>

6 0

3 years ago