PLEASE HELP

Consider a space shuttle which has a mass of about 1.0 x 105 kg and circles the Earth at an altitude of about 200.0 km. Calculat

svetlana [45]

Answer:

1.6675×10^-16N

Explanation:

The force of gravity that the space shuttle experiences is expressed as;

g = GM/r²

G is the gravitational constant

M is the mass = 1.0 x 10^5 kg

r is the altitude = 200km = 200,000m

Substitute into the formula

g = 6.67×10^-11 × 1.0×10^5/(2×10^5)²

g = 6.67×10^-6/4×10^10

g = 1.6675×10^{-6-10}

g = 1.6675×10^-16N

Hence the force of gravity experienced by the shuttle is 1.6675×10^-16N

6 0

2 years ago

Suppose that a wind is blowing in the direction S45°E at a speed of 30 km/h. A pilot is steering a plane in the direction N60°E

Kay [80]

Answer:

The true course: $40.29^\circ$ north of east

The ground speed of the plane: 96.68 m/s

Explanation:

Given:

$V_w$ = velocity of wind = $30\ km/h\ S45^\circ E = (30\cos 45^\circ\ \hat{i}-30\sin 45^\circ\ \hat{j})\ km/h = (21.21\ \hat{i}-21.21\ \hat{j})\ km/h$

$V_p$ = velocity of plane in still air = $100\ km/h\ N60^\circ E = (100\cos 60^\circ\ \hat{i}+100\sin 60^\circ\ \hat{j})\ km/h = (50\ \hat{i}+86.60\ \hat{j})\ km/h$

Assume:

$V_r$ = resultant velocity of the plane

$\theta$ = direction of the plane with the east

Since the resultant is the vector addition of all the vectors. So, the resultant velocity of the plane will be the vector sum of the wind velocity and the plane velocity in still air.

$\therefore V_r = V_p+V_w\\\Rightarrow V_r = (50\ \hat{i}+86.60\ \hat{j})\ km/h+(21.21\ \hat{i}-21.21\ \hat{j})\ km/h\\\Rightarrow V_r = (71.21\ \hat{i}+65.39\ \hat{j})\ km/h$

Let us find the direction of this resultant velocity with respect to east direction:

$\theta = \tan^{-1}(\dfrac{65.39}{71.21})\\\Rightarrow \theta = 40.29^\circ$

This means the the true course of the plane is in the direction of $40.29^\circ$ north of east.

The ground speed will be the magnitude of the resultant velocity of the plane.

$\therefore Magnitude = \sqrt{71.21^2+65.39^2} = 96.68\ km/h$

Hence, the ground speed of the plane is 96.68 km/h.

5 0

3 years ago

A 1400 kg car is traveling east on the highway at 31 m/s and collides into the rear of a slower moving pickup truck of 2400 kg,

zloy xaker [14]

Answer: 31 m/s due east

Explanation: this question can be solved using the law of conservation of linear momentum.

This law states that in a closed or isolated system, during collision, the vector sum of momentum before collision equals the vector sum of momentum after collision.

Momentum = mass × velocity

From our question, our parameters before collision are given below as

Mass of car = mc = 1400kg

Speed of car =vc = 31 m/s (due east)

Mass of truck = mt = 2400kg

Velocity of truck = vt = 25 m/s ( due east )

After collision

Velocity of car = ?

Velocity of truck = 34 m/s ( due east )

Vector sum of momentum before collision is given as

1400 (31) + 2400 (25) = 43400 + 60000 = 103400 kgm/s

After collision the truck is seen to move faster (v = 34 m/s) which implies that the car also moves due east .

1400 (v) + 2400(25) .... A positive value is between both momenta because they are in the same direction.

After collision, we have that

1400v + 60000

Vector sum of momentum before collision = vector sum of momentum after collision

103400 = 1400v + 60000

103400 - 60000 = 1400v

43400 = 1400v

v = 43400/ 1400

v = 31 m/s due east

4 0

3 years ago

The minute hand of a wall clock measures 16 cm from its tip to the axis about which it rotates. The magnitude and angle of the d

olya-2409 [2.1K]

Answer:

Explanation:

Given

Minute hand length =16 cm

Time at a quarter after the hour to half past i.e. 1 hr 45 min

Angle covered by minute hand in 1 hr is 360 and in 45 minutes 270

$|r|=\frac{3\times 2\pi r}{4}=75.408 cm$

$Angle =270^{\circ}$

(c)For the next half hour

Effectively it has covered 2 revolution and a quarter

$|r|=\frac{2\pi r}{4}=25.136 cm$

angle turned $=90^{\circ}$

(f)Hour after that

After an hour it again comes back to its original position thus displacement is same =25.136

Angle turned will also be same i.e. $90 ^{\circ}$

7 0

3 years ago

Name three types of energy that exist in a large piece of charcoal on a grill in the sunlight. Explain why the charcoal has each

qaws [65]

Answer:

A large piece of charcoal on a grill in the sunlight (if it's burning) will consist of the following types of energy:

Chemical
Heat and
Light

Explanation:

Charcoal is basically carbon which is produced when wood is heated strongly in the absence of oxygen. From a chemistry point of view, charcoal contains combustible carbon whose chemical formula is C. Sometimes, $H_{2}O$ which is water may be found in it but in very small units.

All matter contains Heat energy. Charcoal is not an exception. As the charcoal burns, the heat energy is produced along with Light energy.

Light comes in many forms such as Infrared rays, Xrays, Visible Spectrum light, etc.

The glow which the coal gives off fall under the visible spectrum of light.

Cheers

8 0

2 years ago