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3 years ago

What does altered mean

Physics

2 answers:

vladimir1956 [14]3 years ago

8 0

Change or cause to change in character or composition, typically in a comparatively small but significant way.

Elina [12.6K]3 years ago

5 0

Altered means to change something.

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escribe the differences between the terrestrial and jovian planets and classify each of the 8 planets to the proper type

Radda [10]

Answer:

*******

Explanation:

is obviously the correct answer they are both so different yet so alike

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3 years ago

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What is the net force on a car if the force of friction is 15 N and the forward force due to the engine is 20 N?

anygoal [31]

D. 35n forwards....................

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3 years ago

A ball is thrown horizontally from the top of a building 0.10 km high. The ball strikes the ground at a point 65 m horizontally

aivan3 [116]

Answer:

option B

Explanation:

given,

height of building = 0.1 km

ball strikes horizontally to ground at = 65 m

speed at which the ball strike = ?

vertical velocity = 0 m/s

time at which the ball strike

$s = \dfrac{1}{2}gt^2$

$t = \sqrt{\dfrac{2s}{g}}$

$t = \sqrt{\dfrac{2\times 100}{9.8}}$

t = 4.53 s

vertical velocity at the time 4.53 s = g × t = 9.8 × 4.53 = 44.39 m/s

horizontal velocity = $\dfrac{65}{4.53}$ =14.35 m/s

speed of the ball = $\sqrt{44.39^2+14.35^2}$

= 46.65 m/s

hence, the speed of the ball just before it strike the ground = 47 m/s

The correct answer is option B

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3 years ago

A dart is thrown at a dartboard 3.66 m away. When the dart is released at the same height as the center of the dartboard, it hit

lapo4ka [179]

Answer:

The angle is $\theta = 15.48^o$

Explanation:

From the question we are told that

The distance of the dartboard from the dart is $d = 3.66 \ m$

The time taken is $t = 0.455 \ s$

The horizontal component of the speed of the dart is mathematically represented as

$u_x = ucos \theta$

where u is the the velocity at dart is lunched

$distance = velocity \ in \ the\ x-direction * time$

substituting values

$3.66 = ucos \theta * (0.455)$

=> $ucos \theta = 8.04 \ m/s$

From projectile kinematics the time taken by the dart can be mathematically represented as

$t = \frac{2usin \theta }{g}$

=> $usin \theta = \frac{g * t}{2 }$

$usin \theta = \frac{9.8 * 0.455}{2 }$

$usin \theta = 2.23$

=> $tan \theta = \frac{usin\theta }{ucos \theta } = \frac{2.23}{8.04}$

$\theta = tan^{-1} [0.277]$

$\theta = 15.48^o$

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3 years ago

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3 years ago