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3 years ago

A spacecraft is launched from Earth toward the moon.

Physics

2 answers:

elena55 [62]3 years ago

6 0

Answer:

closer to the moon

Explanation:

got it right on time 4 learning

Leona [35]3 years ago

5 0

Closer to the moon than to Earth because Earth has a greater mass than the moon

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The voltage across the terminals of a 9.0 v battery is 8.5 v when the battery is connected to a 60 ω load. part a what is the ba

snow_lady [41]

Refer to the diagram shown below.

i = the current in the circuit., A
R₁ = the internal resistance of the battery, Ω
R₂ = the resistance of the 60 W load, Ω

Because the resistance across the battery is 8.5 V instead of 9.0 V, therefore
(R₁ )(i A) = 9 - 8.5 = (0.5 V)
R₁*i = 0.5 (10

Also,
R₂*i = 9.5 (2)

Because the power dissipated by R₂ is 60 W, therefore
i²R₂ = 60
From (2), obtain
i*9.5 = 60
i = 6.3158 A

From (1), obtain
6.3158*R₁ = 0.5
R₁ = 0.5/6.3158 = 0.0792 Ω = 0.08 Ω (nearest hundredth)

Answer: 0.08 Ω

3 0

3 years ago

Read 2 more answers

A hammer falls from a construction worker's hand 30m above the ground at the same time that a filled soft drink can is projected

drek231 [11]

For harmer:-

initial velocity=u=0m/s

Acceleration due to gravity=10m/s^2

Now

$\\ \sf\longmapsto s=ut+\dfrac{1}{2}gt^2$

$\\ \sf\longmapsto s=0t+\dfrac{1}{2}(10)t^2$

$\\ \sf\longmapsto s=5t^2\dots(1)$

For soft drink:-

u=24m/s

$\\ \sf\longmapsto 30-s=(24)t+\dfrac{1}{2}(10)t^2$

Using eq(1)

$\\ \sf\longmapsto 30-5t^2=24t-5t^2$

$\\ \sf\longmapsto 24t=30$

$\\ \sf\longmapsto t=\dfrac{30}{24}$

$\\ \sf\longmapsto t=1.2s$

After 1.2s they will meet.

Now

putting t in eq(1)

$\\ \sf\longmapsto s=5t^2=5(1.2)^2=5(1.44)=7.2m$

At the height of 7.2m they will meet

3 0

3 years ago

Can someone please shows me the steps and answer. Urgent!

BabaBlast [244]

Using the formula t=root of 2h/g then where h=28 and g=9.8 then substitute so the answer is 2.4seconds

6 0

3 years ago

What are 3 types of contact forces

Svetach [21]

There are different types of contact forces like normal Force, spring force, applied force and tension force.

5 0

3 years ago

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A luggage handler pulls a suitcase of mass 19.6 kg up a ramp inclined at an angle 24.0 ∘ above the horizontal by a force F⃗ of m

Dvinal [7]

(a) 638.4 J

The work done by a force is given by

$W=Fd cos \theta$

where

F is the magnitude of the force

d is the displacement of the object

$\theta$ is the angle between the direction of the force and the displacement

Here we want to calculate the work done by the force F, of magnitude

F = 152 N

The displacement of the suitcase is

d = 4.20 m along the ramp

And the force is parallel to the displacement, so $\theta=0^{\circ}$ . Therefore, the work done by this force is

$W_F=(152)(4.2)(cos 0)=638.4 J$

b) -328.2 J

The magnitude of the gravitational force is

W = mg

where

m = 19.6 kg is the mass of the suitcase

$g=9.8 m/s^2$ is the acceleration of gravity

Substituting,

$W=(19.6)(9.8)=192.1 N$

Again, the displacement is

d = 4.20 m

The gravitational force acts vertically downward, so the angle between the displacement and the force is

$\theta= 90^{\circ} - \alpha = 90+24=114^{\circ}$

Where $\alpha = 24^{\circ}$ is the angle between the incline and the horizontal.

Therefore, the work done by gravity is

$W_g=(192.1)(4.20)(cos 114^{\circ})=-328.2 J$

c) 0

The magnitude of the normal force is equal to the component of the weight perpendicular to the ramp, therefore:

$R=mg cos \alpha$

And substituting

m = 19.6 kg

g = 9.8 m/s^2

$\alpha=24^{\circ}$

We find

$R=(19.6)(9.8)(cos 24)=175.5 N$

Now: the angle between the direction of the normal force and the displacement of the suitcase is 90 degrees:

$\theta=90^{\circ}$

Therefore, the work done by the normal force is

$W_R=R d cos \theta =(175.4)(4.20)(cos 90)=0$

d) -194.5 J

The magnitude of the force of friction is

$F_f = \mu R$

where

$\mu = 0.264$ is the coefficient of kinetic friction

R = 175.5 N is the normal force

Substituting,

$F_f = (0.264)(175.5)=46.3 N$

The displacement is still

d = 4.20 m

And the friction force points down along the slope, so the angle between the friction and the displacement is

$\theta=180^{\circ}$

Therefore, the work done by friction is

$W_f = F_f d cos \theta =(46.3)(4.20)(cos 180)=-194.5 J$

e) 115.7 J

The total work done on the suitcase is simply equal to the sum of the work done by each force,therefore:

$W=W_F + W_g + W_R +W_f = 638.4 +(-328.2)+0+(-194.5)=115.7 J$

f) 3.3 m/s

First of all, we have to find the work done by each force on the suitcase while it has travelled a distance of

d = 3.80 m

Using the same procedure as in part a-d, we find:

$W_F=(152)(3.80)(cos 0)=577.6 J$

$W_g=(192.1)(3.80)(cos 114^{\circ})=-296.9 J$

$W_R=(175.4)(3.80)(cos 90)=0$

$W_f =(46.3)(3.80)(cos 180)=-175.9 J$

So the total work done is

$W=577.6+(-296.9)+0+(-175.9)=104.8 J$

Now we can use the work-energy theorem to find the final speed of the suitcase: in fact, the total work done is equal to the gain in kinetic energy of the suitcase, therefore

$W=\Delta K = K_f - K_i\\W=\frac{1}{2}mv^2\\v=\sqrt{\frac{2W}{m}}=\sqrt{\frac{2(104.8)}{19.6}}=3.3 m/s$

6 0

3 years ago