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2 years ago

What is the speed of a shark that traveled 15 miles in 1 hour

Physics

2 answers:

Elodia [21]2 years ago

6 0

Soln_

T = 1 hrs × 15 mils

=60 × 15

=900 ans#

Assoli18 [71]2 years ago

3 0

900 ex: I have no clue what I’m talking ab I just need points

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Ariel's puppy weighs 15 lb. Heather's puppy weighs 50 N. Whose puppy weighs more?

yarga [219]

15 lb is nearly 6 kg amd 50 n is nearly 5 kg,so first puppy weighs more

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2 years ago

which actions most likely cause the domains within a material to lose their alignment and become more randomized

zimovet [89]

answer:

heating the material

placing the material in a magnetic field of opposite polarity

hitting the material

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2 years ago

Read 2 more answers

An ice skater is spinning at 6.00 rev/s with his moment of inertia being 0.400 kg/m2. Calculate his new moment of inertia if he

labwork [276]

Answer:

New moment of inertia will be $I=1.92kgm^2$

Explanation:

It is given initially angular velocity $\omega =6rev/sec=6\times 2\pi =37.68rad/sec$

Moment of inertia $I=0.4kgm^2$

Angular momentum is equal to $L=I\omega =37.68\times 0.4=15.072kgm^2/sec$

Now angular velocity is decreases to $\omega =1.25rev/sec=1.25\times 2\times 3.14=7.85rad/sec$

As we know that angular momentum is conserved

So $15.072=I\times 7.85$

$I=1.92kgm^2$

So new moment of inertia will be $I=1.92kgm^2$

4 0

3 years ago

A 100-W (watt) light bulb has resistance R=143Ω (ohms) when attached to household current, where voltage varies as V=V0sin(2πft)

Phantasy [73]

Complete Question

A 100-W (watt) light bulb has resistance R=143Ω (ohms) when attached to household current, where voltage varies as V=V0sin(2πft), where V0=110 V, f=60 Hz. The power supplied to the bulb is P=V2R J/s (joules per second) and the total energy expended over a time period [0,T] (in seconds) is $U = \int\limits^T_0 {P(t)} \, dt$

Compute U if the bulb remains on for 5h

Answer:

The value is $U = 7.563 *10^{5} \ J$

Explanation:

From the question we are told that

The power rating of the bulb is $P = 100 \ W$

The resistance is $R = 143 \ \Omega$

The voltage is $V = V_o sin [2 \pi ft]$

The energy expanded is $U = \int\limits^T_0 {P(t)} \, dt$

The voltage $V_o = 110 \ V$

The frequency is $f = 60 \ Hz$

The time considered is $t = 5 \ h = 18000 \ s$

Generally power is mathematically represented as

$P = \frac{V^2}{ R}$

=> $P = \frac{( 110 sin [2 \pi * 60t])^2}{ 144}$

=> $P = \frac{ 110^2 [ sin [120 \pi t])^2}{ 144}$

$U = \int\limits^T_0 { \frac{ 110^2* [sin [120 \pi t])^2}{ 144}} \, dt$

=> $U = \frac{110^2}{144} \int\limits^T_0 { ( sin^2 [120 \pi t]} \, dt$

=> $U = \frac{110^2}{144} \int\limits^T_0 { \frac{1 - cos 2 (120\pi t)}{2} } \, dt$

=> $U = \frac{110^2}{144} \int\limits^T_0 { \frac{1 - cos 240 \pi t)}{2} } \, dt$

=> $U = \frac{110^2}{144} [\frac{t}{2} - [\frac{1}{2} * \frac{sin(240 \pi t)}{240 \pi} ] ]\left | T} \atop {0}} \right.$

=> $U = \frac{110^2}{144} [\frac{t}{2} - [\frac{1}{2} * \frac{sin(240 \pi t)}{240 \pi} ] ]\left | 18000} \atop {0}} \right.$

$U = \frac{110^2}{144} [\frac{18000}{2} - [\frac{1}{2} * \frac{sin(240 \pi (18000))}{240 \pi} ] ]$

=> $U = 7.563 *10^{5} \ J$

7 0

2 years ago

In terms of saturation (unsaturated, saturated, super-saturated). How would you classify the following?

miskamm [114]

Answer:

1. A NaCl solution with a concentration of 50g/100mL of water at 40°C: The NaCl solution with a given concentration is saturated at this temperature .As the temperature increases the solution will more dissolves.

2. A sugar solution with a concentration of 200g/100mL of water at 40°C: The sugar solution with a given concentration is saturated at this temperature. As the temperature increases the solution will more dissolves.

3. A sugar solution with a concentration of 240g/100mL of water at 40°C: The sugar solution with a given concentration is saturated at given temperature.

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2 years ago