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olchik [2.2K]
3 years ago
7

a car advertisement states that a certain car can accelerate from rest to 70km/h in 7 seconds. find the cars acceleration

Physics
1 answer:
ELEN [110]3 years ago
4 0
<span>a = (v2 - v1)/t= acceleration formula

a = (70 - 0)/7
a = 10 km/hr/sec
-----
It's better to use as few units as possible.
10 km/hr = 10 km*1000 m/km/(1 hr*3600 sec/hr) = 25/9 m/sec
a= 25/9 m/sec/se</span>
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100 points!!! Please help!!!
Greeley [361]

Answer:

41°

Explanation:

Kinetic energy at bottom = potential energy at top

½ mv² = mgh

½ v² = gh

h = v²/(2g)

h = (2.4 m/s)² / (2 × 9.8 m/s²)

h = 0.294 m

The pendulum rises to a height of  above the bottom.  To determine the angle, we need to use trigonometry (see attached diagram).

L − h = L cos θ

cos θ = (L − h) / L

cos θ = (1.2 − 0.294) / 1.2

θ = 41.0°

Rounded to two significant figures, the pendulum makes a maximum angle of 41° with the vertical.

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3 years ago
When air is heated its density________
Trava [24]

Answer:

B

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4 0
2 years ago
A ball is attached to one end of a wire, the other end being fastened to the ceiling. The wire is held horizontal, and the ball
Aleksandr [31]

Answer:

the velocity (magnitude and direction) of the ball)  just before the collision is v_i = 5.144 \ m/s

The  velocity (magnitude and direction) of the ball)  just after  the collision is v_f = - 1.129  \ m/s

Explanation:

According to the law of conservation of energy;

m__{ball}} gh = \frac{1}{2}m__{ball}}v_i^2\\\\2* m__{ball}} gh = m__{ball}}v_i^2\\\\v_i^2 = \frac{2*m_{ball}gh}{m_{ball}}\\\\v_i^2 = 2gh\\\\v_i = \sqrt{2gh} \\\\v_i = \sqrt{2*9.8*1.35}\\\\

v_i = 5.144 \ m/s

Thus; the velocity (magnitude and direction) of the ball)  just before the collision is v_i = 5.144 \ m/s

Since, Air resistance is negligible, and the collision is elastic.

The equation for the conservation of momentum and energy can be expressed as:

v_f = [\frac{m_1 -m_2}{m_1+m_2}]v_i\\\\v_f =  [\frac{m_{ball} -m_{block}}{m_{ball}+m_{block}}]v_i\\\\v_f = [\frac{1.6 -2.5}{1.6+2.5}]*5.144\\\\

v_f = - 1.129  \ m/s

The  velocity (magnitude and direction) of the ball)  just after  the collision is v_f = - 1.129  \ m/s

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2 years ago
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Paul [167]
A) Gravity
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