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Gnom [1K]
3 years ago
7

A hiker walks 11 km due north from camp and then turns and walks 11 km due east. What is the total distance walked by the hiker?

Physics
2 answers:
Nitella [24]3 years ago
5 0
22 km it’s not about how far he traveled from his original starting spot just how far he walked so..22km
AnnyKZ [126]3 years ago
3 0

Answer:

22 km

Explanation:

11km + 11km= 22km

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Increase in Space Suit Pressure 0.0/3.0 points (graded) If the pressure in a space suit increases, how will each of the followin
Art [367]

Answer:

Flexibility Increases

Pre-breathe time decreases

Mass of suit decreases.

Explanation:

Spacesuits are designed for space shuttles when a person goes to explore the galaxy. The spacesuits shuttle era are pressurized at 4.3 pounds per inch. The gas in the suit is 100% of oxygen and there is more oxygen to breathe when the altitude of 10,000 is reached. This will decrease the breathing time and mass of suit.

3 0
3 years ago
Electric fields up to 2.00 × 10 5 N/C have been measured inside of clouds during electrical storms. Neglect the drag force due t
katrin2010 [14]

Answer:

1.9161676647\times 10^{13}\ m/s^2 or 1.9532799844\times 10^{12}g

23.4843749996 m

Yes

Explanation:

E = Electric field = 2\times 10^5\ N/C

c = Speed of light = 3\times 10^8\ m/s

m = Mass of proton= 1.67\times 10^{-27}\ kg

q = Charge of electron = 1.6\times 10^{-19}\ C

Acceleration is given by

a=\dfrac{Eq}{m}\\\Rightarrow a=\dfrac{2\times 10^5\times 1.6\times 10^{-19}}{1.67\times 10^{-27}}\\\Rightarrow a=1.9161676647\times 10^{13}\ m/s^2

Dividing by g

\dfrac{a}{g}=\dfrac{1.9161676647\times 10^{13}}{9.81}\\\Rightarrow a=1.9532799844\times 10^{12}g

The acceleration is 1.9161676647\times 10^{13}\ m/s^2 or 1.9532799844\times 10^{12}g

v^2-u^2=2as\\\Rightarrow s=\dfrac{v^2-u^2}{2a}\\\Rightarrow s=\dfrac{(0.1\times 3\times 10^8)^2-0^2}{2\times 1.9161676647\times 10^{13}}\\\Rightarrow s=23.4843749996\ m

The distance is 23.4843749996 m

The gravitational field is very small compared to the electric field so the effects of gravity can be ignored.

5 0
3 years ago
A car is traveling with a velocity of 24.4 m/s. It accelerates at a constant rate of 3.2m/s2. If the acceleration lasts for 5.6
larisa86 [58]

Answer: <em>4</em><em>2</em><em>.</em><em>3</em><em>2</em><em> </em><em>ms-1</em>

Explanation:

v = u+ at

v = 24.4 + ( 3.2×5.6)

v = 42.32 ms-1

6 0
3 years ago
A projectile is fired from the ground (you can assume the initial height is the same as the ground) in a field so there are no o
Cerrena [4.2K]

Answer:

A) 112 m. B) 27.2 m C) 41.1 m/s i + 13.4 m/s j  D) 43.2 m/s

Explanation:

A) Once fired, no external forces act on the projectile in the x-direction, so it keeps moving to the right at constant speed, which is the projection on the x-axis of  the initial velocity vector:

v₀ₓ = v₀* cos 33º = 49 m/s* cos 33º = 41.1 m/s

In the y-direction, the component of the velocity can be found as the projection of v₀ on the y-axis, as follows:

v₀y = v₀* sin 33º = 49 m/s* sin 33º = 26.7 m/s

Both velocities are independent each other, as no one has a projection on the other.

In the vertical direction, the  projectile is in free fall all time, under the influence of gravity , which accelerates it downward.

So, at any time, in the vertical direction, the velocity can be calculated as follows:

vfy = v₀y -g*t (same equation as for an object thrown upwards)

When the object is at its maximum height, the velocity, in the vertical direction, will be momentarily zero, so we can find the time when this happens as follows:

vfy= 0 ⇒ v₀y = g*t ⇒ t = v₀y / g = 26.7 m/s / 9.8 m/s² = 2.72 s

As the time is the same for both movements, we can replace this value in the expression for the displacement x at constant speed, as follows:

x = v₀ₓ* t = 41.1 m/s* 2.72 s = 112 m

B) Like above, as the time is the same for both movements, we can find the time for the instant that the projectile hit the wall, as follows:

x = v₀ₓ* t ⇒ 55. 8 m = 41.1 m/s * t

⇒ t = 55. 8 m / 41.1 m/s = 1.36 s

We can replace this value of t in the equation for the vertical displacement, as follows:

Δy = v₀y*t -1/2*g*t² = (26.7m/s*1.36s) - 1/2*9.8m/s²*(1.36s)² = 27.2 m

C) The velocity of the projectile, at any time, has two components, one horizontal and one vertical.

As explained above, x-component is constant, equal to v₀x:

vx = v₀x i = 41.1 m/s i

For vy, we can apply acceleration definition, using the value of v₀y and t that we have just found:

vfy = voy - g*t = 26.7 m/s - 9.8m/s*1.36 sec = 13.4 m/s

vfy = 13.4 m/s j

v = 41.1 m/s i + 13.4 m/s j

D) Finally, in order to get the speed of the projectile when it hit the wall, we need just to find the magnitude of the velocity, as we get the magnitude of any vector given its vertical and horizontal components:

v = √(41.1 m/s)² +(13.4 m/s)² =43.2 m/s

5 0
3 years ago
A car is stopped at a traffic light. It then travels along a straight road so that its distance from the light is given by x(t)=
True [87]

Answer:

(a) average velocity = 17.6 m/s

(b) when t = 0, v = 0

    when t = 4, v = 19.2 m/s

    when t = 8, v = 28.8 m/s

(c) after starting from rest, the car will be at rest again in 20 s

Explanation:

Given;

x(t)=bt²−ct³, substitute the given values and the equation will become;

x(t)=3t²−0.1t³

(a)average velocity = total distance / total time

total distance, x(t) = 3t²−0.1t³

x(8) = 3t²−0.1t³

X(8) = 3(8)² - 0.1(8)³

X(8) = 140.8 m

total time = 8 s

average velocity = 140.8 / 8

average velocity = 17.6 m/s

(b) instantaneous velocity = dx / dt

dx / dt = 6t - 0.3t²

when t = 0

v = 0

when t = 4 s

v = 6(4) - 0.3(4²) = 19.2 m/s

when t = 8 s

v = 6(8) - 0.3(8²) = 28.8 m/s

(c) the velocity is zero at dx / dt = 0

6t - 0.3t² = 0

t(6 - 0.3t) = 0

t = 0    or  6 - 0.3t = 0

t = 0     or   0.3t = 6

t = 0      or   t = 6 / 0.3

t= 0       or    t =  20 s

After starting from rest, the car will be at rest again in 20 s

6 0
3 years ago
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