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Fynjy0 [20]
3 years ago
6

An elevator weighing 15,000 newtons is raised one story in 50 seconds. If the distance between stories is 3.0 meters, how much p

ower is required of the motor? 750,000 W 45,000 W 5,000 W 900 W
Mathematics
1 answer:
Dennis_Churaev [7]3 years ago
4 0

Answer:

900 W

Step-by-step explanation:

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how many leaves on a tree diagram are needed to represent all possible combinations tossing a coin 3 times
Lelechka [254]
I believe you would do 2 times 3, which gives you 6. 3 is for the number of coin tosses, and 2 is the number of sides of a coin.
6 0
3 years ago
A new DVD is available for sale in a store one week after its release. The cumulative revenue, $R, from sales of the DVD in this
AnnyKZ [126]

Solution :

It is given that :

$f'(t) = (350 \ln   t)'$

       $=350(\ln t)'$

        $=\frac{350}{t}$

So, f(5)=350 \ln (5) \approx 563

     $f'(5) = \frac{350}{5}$

              =70

The relative change is then,

$\frac{f'(5)}{f(5)}=\frac{70}{350\ \ln(5)}$

         $=\frac{1}{5\ \ln(5)}$

         $\approx 0.12$

          =12\%

This means that after 5 weeks, the revenue from the DVD sales in $563 with a rate of change of $70 per week and the increasing at a continuous rate of 12% per week.

3 0
3 years ago
10th grade Geo | What does JK equal?
ladessa [460]

Answer:

16

Step-by-step explanation:

6x-7=2x+3 Cause point M is the midpoint, so JM and MK have to be equal.

4x=10

x=2.5

JM=15-7=8 and MK=5+3=8

8+8=16

4 0
3 years ago
A bucket that weighs 6 lb and a rope of negligible weight are used to draw water from a well that is 80 ft deep. The bucket is f
zaharov [31]

Answer:

3200 ft-lb

Step-by-step explanation:

To answer this question, we need to find the force applied by the rope on the bucket at time t

At t=0, the weight of the bucket is 6+36=42 \mathrm{lb}

After t seconds, the weight of the bucket is 42-0.15 t \mathrm{lb}

Since the acceleration of the bucket is the force on the bucket by the rope is equal to the weight of the bucket.

If the upward direction is positive, the displacement after t seconds is x=1.5 t

Since the well is 80 ft deep, the time to pull out the bucket is \frac{80}{2}=40 \mathrm{~s}

We are now ready to calculate the work done by the rope on the bucket.

Since the displacement and the force are in the same direction, we can write

W=\int_{t=0}^{t=36} F d x

Use x=1.5 t and F=42-0.15 t

W=\int_{0}^{36}(42-0.15 t)(1.5 d t)

=\int_{0}^{36} 63-0.225 t d t

=63 \cdot 36-0.2 \cdot 36^{2}-0=3200 \mathrm{ft} \cdot \mathrm{lb}

=\left[63 t-0.2 t^{2}\right]_{0}^{36}

W=3200 \mathrm{ft} \cdot \mathrm{lb}

4 0
3 years ago
Algebraic divide 24y - 4 by 2=
il63 [147K]
Wouldn't it be 12y-2? or is there information you are leaving out?
7 0
3 years ago
Read 2 more answers
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