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Yakvenalex [24]

3 years ago

14

What is the frequency of a wave that has a speed of 5 m/s and a wavelength of 0.5meter?

1 answer:

Debora [2.8K]3 years ago

5 0

Explanation:

frequency =speed/wavelength

=5/0.5=10Hz

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Why are the alkali metals likely to react with group 17 elements?

Rudiy27

Answer:

Because alkali metals are so reactive, they are found in nature only in combination with other elements. They often combine with group 17 elements, which are very “eager” to gain an electron.

Explanation:

hope this helps you if it does please mark brainliest

7 0

3 years ago

A war wolf is a device used during the middle ages to assault fortifications with large rocks. A simple trebuchet is constructed

Katarina [22]

Answer:v=41.23 m/s

Explanation:

Given

mass of heavy object $m_1=52 kg$

distance of $m_1$ from the axle $r_1=14 cm$

mass of rock $m_2=123 gm$

Length of rod $=4.1 m$

distance of $m_2$ from axle $r_2=4.1-0.14=3.96 m$

Net torque acting is

$T_{net}=m_1gr_1-m_2gr_2$

$T_{net}=52\times 0.14\times g-0.123\times 3.96\times g$

$T_{net}=6.793\times 9.8$

$T_{net}=66.57 N-m$

Work done by $T_{net}$ is converted to rock kinetic Energy

thus

$T_{net}\times \theta =\frac{mv^2}{2}$

Where $\theta =angle\ turned =\frac{\pi }{2}$

$v= velocity\ at\ launch$

$66.57\times \frac{\pi }{2}=\frac{0.123\times v^2}{2}$

$v^2=66.57\times \pi$

$v=\sqrt{1700.511}$

$v=41.23 m/s$

3 0

4 years ago

What is the net power needed to change the speed of a 1600-kg sport utility vehicle from 15.0 m/s to 40.0 m/s in 4.00 seconds

Sergio039 [100]

Answer:

The net power needed to change the speed of the vehicle is 275,000 W

Explanation:

Given;

mass of the sport vehicle, m = 1600 kg

initial velocity of the vehicle, u = 15 m/s

final velocity of the vehicle, v = 40 m/s

time of motion, t = 4 s

The force needed to change the speed of the sport vehicle;

$F = \frac{m(v-u)}{t} \\\\F = \frac{1600(40-15)}{4} \\\\F = 10,000 \ N$

The net power needed to change the speed of the vehicle is calculated as;

$P_{net} = \frac{1}{2} F[u + v]\\\\P_{net} = \frac{1}{2} \times 10,000[15 + 40]\\\\P_{net} = 275,000 \ W$

3 0

3 years ago

The radius of the wheel is 40cm and the radius of the

snow_tiger [21]

Answer:

M.A = load/ effort

1200N/400N

= 3

velocity ratio= radius of wheel/radius of Axle

40cm/10cm

=4

efficiency= 3/4*100

75%

4 0

3 years ago

El resultado de pasar a notación científica la cifra 2 000 000 000 · 10 -15

tensa zangetsu [6.8K]

19,999,985 nomas creo y soy positivo que si es pero puedes ablar ingles

6 0

3 years ago