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garri49 [273]
3 years ago
11

A cat is on a 1.5 m high table and jumps at a 37 angle with a speed of 10 m/s. how far horizontally from the edge of the table d

oes the cat land?
Physics
1 answer:
Kaylis [27]3 years ago
8 0
The answer is 11.51 meters (:
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An B flat is played on a french horn, whose frequency is 466Hz.
scoray [572]

Answer:

Period of one vibration = 0.00215 second (Approx.)

Wavelength {Is speed of sound is 343 m/s] = 0.736 m (Approx.)

Explanation:

Given:

Frequency of wave = 466 Hz

Find:

Period of one vibration

Wavelength {Is speed of sound is 343 m/s]

Computation:

Period of one vibration = 1/F

Period of one vibration = 1 / 466

Period of one vibration = 0.00215 second (Approx.)

Wavelength = Velocity / Frequency

Wavelength {Is speed of sound is 343 m/s] = 343 / 466

Wavelength {Is speed of sound is 343 m/s] = 0.736 m (Approx.)

3 0
3 years ago
Read 2 more answers
To meet the productivity goal, you must move 75 units of product per hour. You have moved 30 units of product in 30 minutes. How
9966 [12]

Answer:

1.5 unit of product per min

Explanation:

30 units of product was moved in 30 minutes.

Number of units left = Total number of units-number of units moved

                                =75-30 =45 units

45 units is available to be moved for the rest 30 min. To be able to achieve this goal of 75 units of product per hour.

45/30 amount of units must be moved in 1 min

=1.5 unit per min

4 0
3 years ago
A car moves at a constant speed of 10 m/s. If the car doesn't accelerate during the next 40 s how far will it go?
tangare [24]

the answer is b. space = time * velocity

8 0
3 years ago
Which feature is a physical property of copper? A oven mitts B a wooden spoon C a metal bowl D rubber soled shoes
FrozenT [24]

metal bowel

because copper is metal...

8 0
3 years ago
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A wire 2.80 m in length carries a current of 5.20 A in a region where a uniform magnetic field has a magnitude of 0.430 T. Calcu
galina1969 [7]

Question:

A wire 2.80 m in length carries a current of 5.20 A in a region where a uniform magnetic field has a magnitude of 0.430 T. Calculate the magnitude of the magnetic force on the wire assuming the following angles between the magnetic field and the current.

(a)60 (b)90 (c)120

Answer:

(a)5.42 N (b)6.26 N (c)5.42 N

Explanation:

From the question

Length of wire (L) = 2.80 m

Current in wire (I) = 5.20 A

Magnetic field (B) = 0.430 T

Angle are different in each part.

The magnetic force is given by

F=I \times B \times L \times sin(\theta)

So from data

F = 5.20 A \times 0.430 T \times 2.80 sin(\theta)\\\\F=6.2608 sin(\theta) N

Now sub parts

(a)

\theta=60^{o}\\\\Force = 6.2608 sin(60^{o}) N\\\\Force = 5.42 N

(b)

\theta=90^{o}\\\\Force = 6.2608 sin(90^{o}) N\\\\Force = 6.26 N

(c)

\theta=120^{o}\\\\Force = 6.2608 sin(120^{o}) N\\\\Force = 5.42 N

3 0
3 years ago
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