Please help with any of them!!!!!!

A sailboat is traveling across the ocean with a speed of 4 m/s when large gusts of wind pick up that starts to accelerate the sa

MatroZZZ [7]

Initial velocity, u = 4 m/s

acceleration due to gusts of wind = 3 m/s^2

time, t = 1 min = 60 s

Let distance travelled = S

From equation of motion,

$S = ut + \frac{1}{2} a {t}^{2} \\ \\ S = 4 \times 60 + \frac{1}{2} \times 3 \times {60}^{2} \\ \\ S = 240 + 5400 \\ \\ S = 5640 \: m$

Thus, the boat would have traveled 5640m after gusts picked up.

4 0

3 years ago

Read 2 more answers

How would you write the number 6,500,000,000 in scientific notation?

photoshop1234 [79]

Scientific form = 6.5 x 109.

8 0

3 years ago

Long Jump: inital center of mass height of 1.08 m, final center of mass height of 0.42 m, projection velocity of 8.7 m/s, projec

sammy [17]

Answer:

1) The maximum jump height is reached at A. $0.337s$

2) The maximum center of mass height off of the ground is B. $1.64m$

3) The time of flight is C. $0.834s$

4) The distance of jump is B. $7.49m$

Explanation:

First of all we need to decompose velocity in its rectangular components, so

$v_{xi}=8.7m/s(cos 22.3\°)=8.05m/s= constant\\v_{yi}=8.7m/s(sin 22.3\°)=3.3m/s$

1) We use, $v_{fy}=v_{iy}-gt$ , as we clear it for $t$ and using the fact that $v_{fy}=0$ at max height, we obtain $t=\frac{v_{iy}}{g} =\frac{3.3m/s}{9,8m/s^{2}} =0.337s$

2) We can use the formula $y_{max}=y_{i}+v_{iy}t-\frac{gt^{2}}{2}$ for $t=0.337s$ , so

$y_{max}=1.08m+(3.3m/s)(0.337s)-\frac{(9.8m/s^{2})(0.337)^{2}}{2}=1.64m$

3) We can use the formula $y_{f}=y_{i}+v_{iy}t-\frac{gt^{2}}{2}$ , to find total time of fligth, so $0.42=1.08+3.3t-\frac{(9.8)t^{2}}{2}\\0=-4.9t^{2}+3.3t+0.66$ , as it is a second-grade polynomial, we find that its positive root is $t=0.834s$

4) Finally, we use $x=v_{x}t=8.05m/s(0.834s)=6.71m$ , as it has an additional displacement of $0.77m$ due the leg extension we obtain,

$x=6.71m+0.77m=7.48m$ , aprox $x=7.49m$

5 0

3 years ago

A sound wave from a police siren has an intensity of 100.0 W/m² at a certain point; a second sound wave from a nearby ambulance

Irina18 [472]

The sound level of the sound wave due to the ambulance is 140.

<h3>What do you mean by sound?</h3>

In terms of physics, the sound is a vibration that travels through a transmission medium like a gas, liquid, or solid as an acoustic wave. Sound is the reception of these waves and the brain's perception of them in terms of human physiology and psychology. Only acoustic waves with frequencies between roughly 20 Hz and 20 kHz, or the audio frequency range, can cause a human to have an auditory sensation. These are sound waves with wavelengths ranging from 17 meters (56 ft) to 1.7 millimeters in the air at atmospheric pressure (0.67 in). Ultrasounds are sound waves with a frequency higher than 20 kHz that are inaudible to humans. Infrasound refers to sound frequencies below 20 Hz. Animals of different species have different hearing ranges.

To learn more about sound, Visit:

brainly.com/question/9349349

#SPJ4

3 0

1 year ago

supose at 20 degree celsius the resistance of Tungsten thermometer is 154.9. WHen placed in a particular solution , the resistan

saw5 [17]

Answer:

T₂ = 95.56°C

Explanation:

The final resistance of a material after being heated is given by the relation:

R' = R(1 + αΔT)

where,

R' = Final Resistance = 207.4 Ω

R = Initial Resistance = 154.9 Ω

α = Temperature Coefficient of Resistance of Tungsten = 0.0045 °C⁻¹

ΔT = Change in Temperature = ?

Therefore,

207.4 Ω = 154.9 Ω[1 + (0.0045°C⁻¹)ΔT]

207.4 Ω/154.9 Ω = 1 + (0.0045°C⁻¹)ΔT

1.34 - 1 = (0.0045°C⁻¹)ΔT

ΔT = 0.34/0.0045°C⁻¹

ΔT = 75.56°C

but,

ΔT = Final Temperature - Initial Temperature

ΔT = T₂ - T₁ = T₂ - 20°C

T₂ - 20°C = 75.56°C

T₂ = 75.56°C + 20°C

7 0

3 years ago