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1 year ago

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a man is walking inside a bus which is travelling at 56.2 km/h.if the speed of the man relative to the ground is 55.8 km/h,is he

walking towards the front or the back??

1 answer:

nlexa [21]1 year ago

6 0

Answer:

rawr

Explanation:

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The mass of block B is twice the mass of block A, while the mass of the pulley is equal to the mass of block A. The blocks are l

marshall27 [118]

Answer:

C)the right cord pulls on the pulley with greater force than the left cord

Explanation:

As we can see the figure that B is connected to the right string while A is connected to the left string

Now force equation for B as it will move downwards is given as

$(2m)g - T_B = (2m)a$

similarly for block A which will move upwards we can write the equation as

$T_A - mg = ma$

now we know that pulley is also rotating so the tangential acceleration of the rope at the contact point with pulley must be same as that of acceleration of the blocks

so here pulley will rotate clockwise direction

So tension in the right string must be more than the left string

So correct answer will be

C)the right cord pulls on the pulley with greater force than the left cord

3 0

2 years ago

A rock is thrown upward from the level ground in such a way that the maximum height of its flight is equal to its horizontal ran

marshall27 [118]

Answer

a) For the rock

$\dfrac{v_t^2sin 2\theta}{g} = \dfrac{v_t^2sin^2\theta}{2g}$

$2sin\thetacos\theta = \dfrac{sin^2\theta}{2}$

$2cos\theta = \dfrac{sin\theta}{2}$

$tan\theta = 4$

$\theta = tan^{-1} 4$

$\theta = 76^0$

b) $\theta = 45^0$ for maximum range

$\dfrac{d_{max}}{d}=\dfrac{(v_tcos 45^0)(2v_tsin 45^0)g}{(v_tcos 76^0)(2v_tsin 76^0)g}$

$\dfrac{d_{max}}{d}=\dfrac{0.707\times 0.707)}{0.97\times 0.242}$

$\dfrac{d_{max}}{d}=2.129$

$d_{max}=2.129 d$

c) The value of θ is the same on every planet as g divides out.

5 0

3 years ago

An object thrown vertically upward from the surface of a celestial body at a velocity of 36 m/s reaches a height of sequalsminu

Anarel [89]

Answer:

v = -1.8t+36

20 seconds

360 m

40 seconds

36 m/s

The object speed will increase when it is coming down from its highest height.

Explanation:

$s=-0.9t^2+36t$

Differentiating with respect to time we get

$\frac{ds}{dt}=-1.8t+36\\\Rightarrow v=-1.8t+36$

a) Velocity of the object after t seconds is v = -1.8t+36

At the highest point v will be 0

$0=-1.8t+36\\\Rightarrow t=\frac{-36}{-1.8}\\\Rightarrow t=20\ s$

b) The object will reach the highest point after 20 seconds

$s=-0.9t^2+36t\\\Rightarrow s=-0.9\times 20^2+36\times 20\\\Rightarrow s=360\ m$

c) Highest point the object will reach is 360 m

$s=-0.9t^2+36t\\\Rightarrow 360=-0.9t^2+36t\\\Rightarrow -0.9t^2+36t-360=0\\\Rightarrow -9t^2+360t-3600=0$

$\frac{-360\pm \sqrt{0}}{2\left(-9\right)}\\\Rightarrow t=20\ s$

d) Time taken to strike the ground would be 20+20 = 40 seconds

$[tex]v=u+at\\\Rightarrow v=0+0.9\times 2\times 20\\\Rightarrow v=36\ m/s$

Acceleration will be taken as positive because the object is going down. Hence, the sign changes. 2 is multiplied because the expression is given in the form of $s=ut+\frac{1}{2}at^2$

e) The velocity with which the object strikes the ground will be 36 m/s

f) The speed will increase when the object has gone up and for 20 seconds and falls down for 20 seconds. The object speed will increase when it is coming down from its highest height.

4 0

2 years ago

Whenever Dylan walks into his room, his roommate loudly blows an air horn, which startles Dylan. Dylan doesn't think this is fun

Sav [38]

Answer:

stimuli

Explanation:

4 0

2 years ago

A stone is dropped from from rest at the top of a mine shaft. It takes 95 seconds for the stone to fall to the bottom of the min

Marianna [84]

Distance of fall from rest,
without air resistance = (1/2) (gravity) (time)²

   = (1/2) (9.8 m/s²) (95 sec)²

   = (4.9 m/s²) (9,025 sec²)

   = 44,222.5 meters .

The depth of the mine shaft is five times the height of Mt. Everest !

6 0

2 years ago