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aev [14]
3 years ago
7

A sailor pulls a boat horizontally along a dock for 10.0 m with an angled rope. He does 766 J of work and has a force of 100.0 N

. What was the angle of the rope?
O 40°
Physics
1 answer:
Angelina_Jolie [31]3 years ago
8 0

Explanation:

W = Fd cos θ

766 J = (100.0 N) (10.0 m) cos θ

cos θ = 0.766

θ = 40.0°

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A galaxy that is a featureless spherical ball of stars would be called a type
satela [25.4K]

Answer:

E0

Explanation:

Yes. The "0" indicates that it is spherical

7 0
2 years ago
car 1 is traveling south at 18 m/s and has a full load, giving it a total mass of 14,650 kg. Car 2 is traveling north at 11 m/s
kvasek [131]

Answer:

v_{4}= 80.92[m/s] (Heading south)

Explanation:

In order to calculate this problem, we must use the linear moment conservation principle, which tells us that the linear moment is conserved before and after the collision. In this way, we can propose an equation for the solution of the unknown.

ΣPbefore = ΣPafter

where:

P = linear momentum [kg*m/s]

Let's take the southward movement as negative and the northward movement as positive.

-(m_{1}*v_{1})+(m_{2}*v_{2})=-(m_{1}*v_{3})+(m_{2}*v_{4})

where:

m₁ = mass of car 1 = 14650 [kg]

v₁ = velocity of car 1 = 18 [m/s]

m₂ = mass of car 2 = 3825 [kg]

v₂ = velocity of car 2 = 11 [m/s]

v₃ = velocity of car 1 after the collison = 6 [m/s]

v₄ = velocity of car 2 after the collision [m/s]

-(14650*18)+(3825*11)=(14650*6)-(3825*v_{4})\\v_{4}=80.92[m/s]

4 0
2 years ago
At noon, ship A is 140 km west of ship B. Ship A is sailing east at 30 km/h and ship B is sailing north at 25 km/h. How fast (in
Luba_88 [7]

Answer:

The rate of change of distance between the two ships is 18.63 km/h

Explanation:

Given;

distance between the two ships, d = 140 km

speed of ship A = 30 km/h

speed of ship B = 25 km/h

between noon (12 pm) to 4 pm = 4 hours

The displacement of ship A at 4pm = 140 km - (30 km/h x 4h) =

140 km - 120 km = 20 km

(the subtraction is because A is moving away from the initial position and the distance between the two ships is decreasing)

The displacement of ship B at 4pm = 25 km/h x 4h = 100 km

Using Pythagoras theorem, the resultant displacement of the two ships at 4pm is calculated as;

r² = a²   +  b²

r² = 20²  +  100²

r = √10,400

r = 101.98 km

The rate of change of this distance is calculated as;

r² = a²   +  b²

r = 101.98 km, a = 20 km, b = 100 km

2r(\frac{dr}{dt} ) = 2a(\frac{da}{dt} )  + 2b(\frac{db}{dt} )\\\\r(\frac{dr}{dt} ) = a(\frac{da}{dt} )  + b(\frac{db}{dt} )\\\\101.98(\frac{dr}{dt} ) = 20(-30 )  + 100(25 )\\\\101.98(\frac{dr}{dt} ) = -600 + 2,500\\\\101.98(\frac{dr}{dt} ) = 1900\\\\\frac{dr}{dt}  = \frac{1900}{101.98} = 18.63 \ km/h

5 0
3 years ago
A star's parallax angle is 0.8. How far away is the star in parsecs? Astronomy
kramer

Distance= 1/arc seconds

1/.8= 1.25 parsecs away

3 0
3 years ago
Read 2 more answers
Q. 5 A bullet of 10 g strikes a sandbag at a speed of 103 ms-1 and gets embedded after travelling 5 cm. Calculate (i) the resist
Xelga [282]

Answer:

(i) the resistive force exerted by the sand on the bullet is - 1 x 10⁵ N

(ii) the time taken by the bullet to come to rest is 1 x 10⁻⁴ s

Explanation:

Given;

mass of bullet, m = 10 g = 0.01 kg

speed of the bullet, v = 10³ m/s

distance traveled by the bullet, d = 5 cm = 0.05 m

(i) the resistive force exerted by the sand on the bullet

the acceleration of the bullet is given by;

v² = u² + 2as

0 = (10³)² + a(2 x 0.05)

-0.1a = (10⁶)

-a = (10⁶) / 0.1

a = -1 x 10⁷ m/s²

The resistive force is given by;

F = ma

F = (0.01)(-1 x 10⁷)

F = - 1 x 10⁵ N

(ii) the time taken by the bullet to come to rest.

v = u + at

0 = 10³ + (-1 x 10⁷ t)

1 x 10⁷ t = 10³

t = 10³  / 10⁷

t = 1 x 10⁻⁴ s

3 0
3 years ago
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