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3 years ago

an astronaut drops a hammer and an egg while standing on the surface of the moon. which one will the surface first?

2 answers:

7 0

The two will fall at the same speed and reach the surface at the same time. This is because the two will experience the same gravitational acceleration on the moon. However, on the earth surface the two will land on the surface of the earth at the same time due to air resistance such that the egg will experience a higher air resistance than the hammer. On, the moon, where there is no noticeable atmosphere there is no air resistance on either object and both fall at the same speed. It is also important to note that their mass doesn't affect their speed.

german3 years ago

4 0

Answer:

They both reach at the same time.

Explanation:

Let the time taken by the hammer and egg be T and t respectively.

initial velocity, u = 0 for both

Distance traveled, h = h for both

acceleration due to gravity = g for both

By using second equation of motion,

S = ut + 0.5 at^2

For hammer

h = 0 + 0.5 x g x T^2 .... (1)

For egg

h = 0 + 0.5 x g x t^2 ....(2)

By observing equation (1) and equation (2), the time taken by the hammer and the egg is same.

They both reach at the same time.

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An airplane accelerates from a speed of 88m/s to a speed of 132 m/s during a 15 second time interval. How far did the airplane t

Gelneren [198K]

Answer:

$1650\:\mathrm{m}$

Explanation:

We can use the following kinematics equations to solve this problem:

$v_f=v_i+at,\\{v_f}^2={v_i}^2+2a\Delta x$ .

Using the first one to solve for acceleration:

$132=88+a(15),\\15a=44,\\a=\frac{44}{15}=2.9\bar{3}\:\mathrm{m/s^2}$ .

Now we can use the second equation to solve for the distance travelled by the airplane:

$132^2=88^2+2\cdot2.9\bar{3}\cdot \Delta x,\\\Delta x= \frac{9680}{2\cdot2.9\bar{3}},\\\Delta x =\fbox{$ 1650\:\mathrm{m}$}$ (three significant figures).

6 0

3 years ago

In a mattress test, you drop a 7.0 kg bowling ball from a height of 1.5 m above a mattress, which as a result compresses 15 cm a

Assoli18 [71]

Answer:

(d) 113.19 J

(e) 113.19 J

(f) 10061 N/m

Explanation:

15 cm = 0.15 m

Let g = 9.8 m/s2

$E_p = mgx = 7*9.8*0.15 = 10.29 J$

(d) By using law of energy conservation with potential energy reference being 0 at the maximum compression point. As the ball falls and come to a stop at the compression point, its potential energy is transferred to elastic energy, which is the work that the mattress does on the ball:

$E_p = E_e$

$E_e = mgh$

where h = 1.5 + 0.15 = 1.65 m is the vertical distance that it falls.

$E_e = 7*9.8*1.65 = 113.19 J$

(e) Before the compression, the potential energy of the mattress is 0. After the compression, the potential energy is 113.19J. So it has increased by 113.19J due to the potential energy transferred from the falling ball.

(f) $E_e = 113.19 = kx^2/2$

$k0.15^2/2 = 113.19$

$k = 10061 N/m$

8 0

3 years ago

Read 2 more answers

A 2000-kg truck is being used to lift a 400-kg boulder B that is on a 50-kg pallet A. Knowing the acceleration of the rear-wheel

anzhelika [568]

Answer:

(a) reaction at each front wheel is 5272N (upward)

(b) force between boulder and pallet is 4124N (compression)

Explanation:

Acceleration of the truck $a_{t$ = 1 m/ $s^{2}$ (to the left)

when the truck moves 1 m to the left, the boulder is B and pallet A are raised 0.5 m, then,

$a_{A}$ = 0.5 m/ $s^{2}$ (upward) , $a_{B}$ = 0.5 m/ $s^{2}$ (upward)

Let T be tension in the cable

pallet and boulder: ∑fy = ∑(fy)eff = 2T- ( $m_{A}$ + $m_{B}$ )g = ( $m_{A}$ + $m_{B}$ ) $a_{B}$

= 2T- (400 + 50)*(9.81 m/ $s^{2}$ ) = (400 + 50)*(0.5 m/ $s^{2}$ )

T = 2320N

Truck: $M_{R}$ = ∑( $M_{R}$ )eff: = $-N_{f}$ (3.4m) + $m_{T}$ (2.0m) - T (0.6m)= $m_{T} a_{T}$ (1.0m)

Nf = (2.0m)(2000 kg)(9.81 m/ $s^{2}$ )/3.4m - (0.6 m)(2320 N)/3.4m + (1.0 m)(2000 kg)(1.0 m/ $s^{2}$ ) = 11541.2N - 409.4N - 588.2N = 10544N

∑fy (upward) = ∑(fy)eff: $N_{f}$ + $N_{R}$ - $m_{T}$ g = 0

10544 + $N_{R}$ - (2000kg)(9.81 m/ $s^{2}$ ) = 0

$N_{R}$ = 9076N

∑fx (to the left) = ∑(fx)eff: $F_{R}$ - T = $m_{T} a_{T}$

$F_{R}$ = 2320N + (2000kg)(9.81 m/ $s^{2}$ ) = 4320N

(a) reaction at each front wheel:

1/2 $N_{f}$ (upward): 1/2 (10544N) = 5272N (upward)

(b) force between boulder and pallet:

∑fy (upward) = ∑(fy)eff: $N_{B}$ + $M_{B}$ g - $m_{B}$ $a_{B}$

$N_{B}$ = (400kg)(9.81 m/ $s^{2}$ ) + (400kg)(0.5 m/ $s^{2}$ ) = 4124N (compression)

3 0

4 years ago

Read 2 more answers

In the geocentric model which motion would occur?

nasty-shy [4]

D) The Sun would revolve around the Earth. 8. In the geocentric model (the Earth at the center of the universe), which motion would occur? B) the Sun in circular orbits C) Earth in slightly elliptical orbits D) Earth in circular orbits 9.

7 0

3 years ago

A truck pulls a caravan at 85 km/h [S]. When they hit a bump in the road the caravan detaches! If the caravan decelerates at a r

Brrunno [24]

The caravan has an initial speed of

85 km/h × (1000 m/km) × (1/3600 h/s) ≈ 23.611 m/s

If it slows down at a constant rate, then it will cover a distance x such that

0² - (23.611 m/s)² = 2 (-1.3 m/s²) x

Solve for x :

x = -(23.611 m/s)² / (2 (-1.3 m/s²)) ≈ 214.417 m ≈ 210 m

5 0

3 years ago