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2 years ago

7

A box with the mass of 74 kg is sitting on the ground and the bottom of the box measures 90 cm by 60 cm. What is the pressure ex

erted by the box on the ground?

1 answer:

Alex Ar [27]2 years ago

6 0

Answer: 112.5

Explanation:

divide the multiply

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A jogger hears a car alarm and decides to investigate. While running toward the car, she hears an alarm frequency of 872.10 Hz.

Nata [24]

Answer:

v = 4.18 m/s

Explanation:

given,

frequency of the alarm = 872.10 Hz

after passing car frequency she hear = 851.10 Hz

Speed of sound = 343 m/s

speed of the jogger = ?

speed of the

$v_f = \dfrac{872.10-851.10}{2}$

$v_f =10.5\ Hz$

v_o = 872.1 - 10.5

$V_0 = 861.6\ Hz$

The speed of jogger

$v = \dfrac{v_1 \times 343}{v_0}-343$

$v = \dfrac{872.1 \times 343}{861.6}-343$

v = 4.18 m/s

5 0

2 years ago

Give an example in which a small force exerts a large torque. give another example in which a large force exerts a small torque.

goldenfox [79]

<span>Since the torque involves the product of force times lever arm, a small force can exert a greater torque than a larger force if the small force has a large enough lever arm.

With a large force exerts a small torque is a gate, hinged in its vertical line (axis). When pushed from a point near to the hinge, a very large amount is needed to open the gate.
</span><span>
</span>

3 0

2 years ago

Read 2 more answers

A 1.90-m-long barbell has a 25.0 kg weight on its left end and a 37.0 kg weight on its right end. if you ignore the weight of th

gregori [183]

Answer: The center of gravity is 1.1338 m away from the left side of the barbell

Explanation:

Length of the barbell = 1.90 m

The distance center of gravity from left = x

Mass on the left side = 25 kg

The distance center of gravity from right = 1.90 - x

Mass on the right side = 37 kg

At the balance point: $m_1x_1=m_2x_2$

$25 kg\times x=37 kg\times (1.90-x)$

$x=1.1338 m$

The center of gravity is 1.1338 m away from the left side of the barbell

7 0

2 years ago

Read 2 more answers

If I connect an inductor (L) to a capacitor (C), I will get an LC oscillator circuit with some natural frequency omega. If I wer

mart [117]

The new natural frequency would be ω/2.

we know that,

$f = \frac{1}{2 pi \sqrt{LC} }$ = ω. -> equation 1

now, when capacitance is quadrupled,

$f' = \frac{1}{2 pi \sqrt{L ( 4C )} }$

$f' = \frac{1}{2 pi (2)\sqrt{LC} }$ . -> equation 2

substituting value of equation 1 in equation 2 , we get,

$f' = \frac{w}{2}$

Hence, the new natural frequency of the circuit is ω/2.

what do you mean by frequency ?

The resonant frequency for a particular circuit is the frequency at which this equality stands true. Where L is the inductance in henries and C is the capacitance in farads, this is the LC circuit's resonant frequency.

Learn more about frequency here:-

brainly.com/question/12530980

#SPJ4

8 0

1 year ago

A 500 kg sack of coal falls vertically onto a 2000 kg railroad flatcar which was initially moving horizontally at 3 m/s. no exte

Zinaida [17]

Since there are no external forces, including friction, act on the flatcar. after the sack rests on the flatcar, we would assume that momentum is conserved. This means that

total momentum of car before collision = total momentum of car after collision.

Recall,

momentum = mass x velocity

From the information given,

mass of car before collision = 2000

velocity of car before collision = 3

Thus,

total momentum of car before collision = 2000 x 3 = 6000

Also,

mass of sack = 500

mass of car and sack after collision = 500 + 2000 = 2500

velocity after collision = v

momentum after collision = 2500 x v = 2500v

Since momentum is conserved, then

6000 = 2500v

v = 6000/2500

v = 2.4

the speed of the flatcar is 2.4 m/s

6 0

4 months ago