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patriot [66]
2 years ago
11

1-A boy rolls a toy car across a floor with a velocity of 3.21 m/s. How long does it take the car to travel a distance of 4.50 m

?
A-0.71s
B-1.40s
C-2.9s
D-14s
2-A girl heads out for a jog and runs at 2.95 m/s, due North, for 3600 s. How far did she run?
A-0.194 x 10^-4
B-1220m
C-5240
D-10620
3-A car is traveling South on I-85. It travels between two exits that are 5.40 km apart in 4.85 minutes. What is the average velocity of the car in m/s?
A-8.42m/s
B-12.8m/s
C-14.9m/s
D-18.6m/s
4-An airplane takes 1.30 hours to travel to an airport north of Atlanta. If the average speed of the plane is 134 m/s, what is the plane's displacement as measured from Atlanta?
A-129,324m=129,000m rounded
B-356,247m=356,000m rounded
C-498,782m=499,000m rounded
D-627,120m=627,000m rounded
5-How long does it take a sailboat traveling 18.0 m/s to go 15.7 km west?
A-15,000s
B-872s
C-594s
D-326s
Physics
1 answer:
Maurinko [17]2 years ago
4 0
B 4.5/3.21 = 1.4
B 3600/2.95 = 1220
D 5.4*1000=1113.4/60=18.6
D 60*1.3=78*60=4,680*134=627,120
B 15.7*1000=15,700/18=872
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A rocket lifts off the pad at cape canaveral. according to newton's law of gravitation, the force of gravity on the rocket is gi
Alenkinab [10]
The equation is the Law of Universal Gravitation. The gravitational constant G is equal to 6.67×10⁻¹¹ Nm²/kg². The mass of the Earth is <span>5.972 ×10</span>²⁴ kg. Compared to the mass of the Earth, the mass of the rocket is negligible. So, we don't need to know the mass of the rocket. Substituting the values:

F = (6.67×10⁻¹¹ Nm²/kg²)(5.972 ×10²⁴ kg)/(4000 miles*(1.609 km/1mile))²
F = 9616423.08 N

The work is equal to
W = Fd
W = (9616423.08 N)(2000 miles*1.609 km/mile)
W = 9.095×10¹⁰ Joules
8 0
2 years ago
A solid block of mass m is suspended in a liquid by a thread. The density of the block is greater than that of the liquid. Initi
Tanzania [10]

Answer:

(C) T

The tension T at equilibrium will be equal to the Buoyant force.

The Buoyant force is given by:

Fb = density x acceleration due to gravity x volume displaced

The change in height doesn't affect the Buoyant force and hence the tension.

Note: The figure of question is added in the attachment

3 0
2 years ago
An 80.0-kg object is falling and experiences a drag force due to air resistance. The magnitude of this drag force depends on its
Alja [10]

Answer:

 Terminal velocity of object = 12.58 m/s

Explanation:

 We know that the terminal velocity is attained when drag force and gravitational force are of the same magnitude.

Gravitational force = mg = 80 * 9.8 = 784 N

Drag force = 12.0v+4.00v^2

Equating both, we have

    784=12.0v+4.00v^2\\ \\ v^2+3v-196=0\\ \\ (v-12.58)(v+15.58)=0

  So v = 12.58 m/s or v = -15.58 m/s ( not possible)

 So terminal velocity of object = 12.58 m/s    

7 0
2 years ago
What voltage is required to move 6A through 20?<br><br>​
Marina CMI [18]

Answer:

120V

Explanation:

Given parameters:

Current  = 6A

Resistance  = 20Ω

Unknown:

Voltage = ?

Solution:

According to ohms law;

           V = IR

Where V is the voltage

            I is the current

            R is the resistance

Now, insert the parameters and solve;

    V  = 6 x 20  = 120V

7 0
2 years ago
Suppose that you lift four boxes individually, each at a constant velocity. The boxes have weights of 3.0 N, 4.0 N, 6.0 N, and 2
Alona [7]

Answer:

The vertical distance of weight 3.0 N = 4 m, vertical distance of weight 4.0 N = 3 m, vertical distance of weight 6.0 N = 2 m, vertical distance of weight 2.0 N = 6 m

Explanation:

Worked : work can be defined as the product of force and distance.

The S.I unit of work is Joules (J).

Mathematically it can be represented as,

W = F×d.................. Equation 1

d = W/F.............................. Equation 2

where W = work, F = force, d = distance.

<em>Given: W = 12 J</em>

(i) for the 3.0 N weight,

using equation 2

d = 12/3

d= 4 m.

(ii) for the 4.0 N weight,

d = 12/4

d = 3 m.

(iii) for the 6.0 N weight,

d = 12/6

d = 2 m.

(iv) for the 2.0 N weight,

d = 12/2

d = 6 m

Therefore vertical distance of weight 3.0 N = 4 m, vertical distance of weight 4.0 N = 3 m, vertical distance of weight 6.0 N = 2 m, vertical distance of weight 2.0 N = 6 m

8 0
2 years ago
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