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Alexxx [7]
2 years ago
7

A stuntman drives a car with a mass of 1500 kg on a drawbridge. The car accelerates with a constant force of 10,000 N. While he

is driving, the drawbridge is raised to an incline of 30°. What is the car's new acceleration on this incline, ignoring the force due to air resistance? (recall that g=9.8 m/s^2)
Physics
1 answer:
klemol [59]2 years ago
5 0
The cars new accelleration is 9,000 N. 

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Prove that..<br>please help<br>​
GaryK [48]

\large{ \boxed{ \bf{ \color{red}{Universal \: law \: of \: gravitation}}}}

Every object in the universe attracts every other object with a force which is proportional to the product of their masses and inversely proportional to the square of the distance between them. The forces along the line joining the centre of the two objects.

❍ Let us consider two masses m1 and m2 line at a separation distance d. Let the force of attraction between the two objects be F.

According to universal law of gravitation,

\large{ \longrightarrow{ \rm{F \propto m_1 m_2}}}

Also,

\large{ \longrightarrow{ \rm{ F \propto  \dfrac{1}{ {d}^{2} } }}}

Combining both, We will get

\large{ \longrightarrow{ \rm{F  \propto  \dfrac{ m_1 m_2}{ {d}^{2}}}}}

Or, We can write it as,

\large{ \longrightarrow{ \rm{F  \propto  \:  G \dfrac{ m_1 m_2}{ {d}^{2} }}}}

Where, G is the constant of proportionality and it is called 'Universal Gravitational constant'.

☯️ Hence, derived !!

<u>━━━━━━━━━━━━━━━━━━━━</u>

8 0
3 years ago
A car accelerates uniformly from rest to a speed of 10m/s. Calculate its acceleration
Maurinko [17]

Answer:

Give four examples of landforms where both the water and the land around it are flat.

Explanation:

8 0
2 years ago
A 0.0780 kg lemming runs off a
kotegsom [21]

Answer:

5.01 J

Explanation:

Info given:

mass (m) = 0.0780kg

height (h) = 5.36m

velocity (v) = 4.84 m/s

gravity (g) = 9.81m/s^2

1. First, solve for Kinetic energy (KE)

KE = 1/2mv^2

1/2(0.0780kg)(4.84m/s)^2 = 0.91 J

so KE = 0.91 J

2. Next, solve for Potential energy (PE)

PE = mgh

(0.0780kg)(9.81m/s^2)(5.36m) = 4.10 J

so PE = 4.10 J

3. Mechanical Energy , E = KE + PE

Plug in values for KE and PE

KE + PE = 0.91J + 4.10 J = 5.01 J

4 0
2 years ago
An electron (q=-1.602×10-19C) is placed .03m away from spherical object with a net charge of -7.2 C.
vovangra [49]

Answer:

Explanation:

electric field at the location of electron

= 9 x 10⁹ x 7.2 / .03²

= 72 x 10¹² N/C

force on electron = electric field x charge on electron

= 72 x 10¹² x 1.6 x 10⁻¹⁹

= 115.2 x 10⁻⁷ N .

C )

work done = charge on electron x potential difference at two points

potential at .03 m

= 9 x 10⁹ x 7.2 / .03

= 2.16 x 10¹² V

potential at .001 m

= 9 x 10⁹ x 7.2 / .001

= 64.8 x 10¹² V

potential difference = (64.8 - 2.16 )x 10¹² V

= 62.64 x 10¹² V  .

work done = 62.64 x 10¹² x 1.6 x 10⁻¹⁹

= 100.224 x 10⁻⁷ J .

D )

There will be no change in the magnitude of force on positron except that the direction of force will be reversed . In case of electron , there will be repulsion and in case of positron , there will be attraction .

Work done in case of electron will be positive and work done in case of positron will be negative .

electric field due to charge will be same in both the cases .

8 0
2 years ago
A ball rolls down an incline with an acceleration of 10 cm/s^2. If it starts with an initial velocity of 0 cm/s and has a velocy
11Alexandr11 [23.1K]
Given a = 10 cm/s²
          u = 0 cm/s
          v = 50 cm/s
we know that 
         v²=u²+2aS
        2500=2×10×S
        2500÷20 = S
        S= 125 cm
The ramp is 125 cm
8 0
3 years ago
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