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3 years ago

How many seconds would it take to drive to Miami if I go 75 mph the whole way? (use 675 miles for the distance)

Physics

1 answer:

Mekhanik [1.2K]3 years ago

3 0

Answer:

32,400 seconds

Explanation:

675 (distance) / 75 mph = 9 hours of driving.

9 hours x 60 minutes = 540 minutes.

540 minutes x 60 seconds = 32,400 seconds

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It takes 50 seconds for Tyler to do 450 Joules of work. How much power did Tyler use?

Tomtit [17]

Answer:

<h2>The answer is 9 W</h2>

Explanation:

The power used by an object can be found by using the formula

$p = \frac{w}{t} \\$

w is the workdone

t is time taken

From the question we have

$p = \frac{450}{50} = \frac{45}{5} \\$

We have the final answer as

Hope this helps you

8 0

3 years ago

What is the first velocity of the car with four washers at

VARVARA [1.3K]

Answer:

0.28

0.56

Explanation:

5 0

3 years ago

Read 2 more answers

As an airplane is taking off at an airport its position is closely monitored by radar. The following three positions are measure

Sergeeva-Olga [200]

Answer: acceleration a = 25m/s^2

Explanation:

Given that:

The plane travels with constant acceleration

x1 = 241.22 m at t1 = 3.70 s

x2 = 297.60 m at t2 = 4.20 s

x3 = 360.23 m at t3 = 4.70 s.

We need to calculate the velocity in the two time intervals.

Interval 1:

Average Velocity v1 = ∆x/∆t = (x2 - x1)/(t2-t1)

v1 = (297.60-241.22)/(4.20-3.70) = 112.76m/s

Interval 2:

Average Velocity v2 = ∆x/∆t = (x3-x2)/(t3-t2)

v2 = (360.23-297.60)/(4.70-4.20)

v2 = 125.26m/s

Acceleration:

Acceleration a = ∆v/∆t

∆v = v2-v1 = 125.26m/s-112.76m/s = 12.5m/s

∆t = change in average time of the two intervals = (t3-t1)/2 = (4.70-3.70)/2 = 0.5s

a = 12.5/0.5 = 25m/s^2

4 0

3 years ago

Ahmad pulls a sled across level snow by 75 N . If the sled moves 15 m , how much work does Ahmad do on the sled?

Nataly_w [17]

Answer:

1125 J

Explanation:

W=F x s

= 75 x 15

=1125

4 0

3 years ago

A 6.75 nC charge is located 1.99 m from a 4.46 nC point charge.

sladkih [1.3K]

Explanation:

Given that,

Charge 1, $q_1=6.75\ nC=6.75 \times 10^{-9}\ C$

Charge 2, $q_2=4.46\ nC=4.46\times 10^{-9}\ C$

The distance between charges, r = 1.99 m

To find,

The electrostatic force and its nature

Solution,

(a) The electric force between two charges is given by :

$F=\dfrac{kq_1q_2}{r^2}$

$F=\dfrac{9\times 10^9\times 6.75\times 10^{-9}\times 4.46\times 10^{-9}}{(1.99)^2}$

$F=6.84\times 10^{-8}\ N$

(b) As the magnitude of both charges is positive, then the force between charges will be repulsive.

Therefore, this is the required solution.

5 0

4 years ago