Travels 10m north followed by 24m East. What is the distance the student travelled in meters

Answer:

the answer is 234 cm is 2.34 meters

Answer:

1.5 m

Explanation:

a = v/t

a = 2/1,5 = 4/3

x = 1/2 a t² + vt + x

x = 1/2 × 4/3 × 9/4 = 3/2 = 1.5 m

Answer:

1.86 m

Explanation:

First, find the time it takes to travel the horizontal distance.  Given:

Δx = 52 m

v₀ = 26 m/s cos 31.5° ≈ 22.2 m/s

a = 0 m/s²

Find: t

Δx = v₀ t + ½ at²

52 m = (22.2 m/s) t + ½ (0 m/s²) t²

t = 2.35 s

Next, find the vertical displacement.  Given:

v₀ = 26 m/s sin 31.5° ≈ 13.6 m/s

a = -9.8 m/s²

t = 2.35 s

Find: Δy

Δy = v₀ t + ½ at²

Δy = (13.6 m/s) (2.35 s) + ½ (-9.8 m/s²) (2.35 s)²

Δy = 4.91 m

The distance between the ball and the crossbar is:

4.91 m − 3.05 m = 1.86 m

Given :

A car drives 10 km with a speed of 72 km/hr and then runs out of gas.

Then you walk 2 km for the next 30 min  until you find a gas station .

To Find :

The average speed of whole journey .

Solution :

Average speed is given by :

[tex]v_{avg}=\dfrac{\text{Total distance covered}}{\text{Total time taken}}\\\\v_{avg}=\dfrac{10+2}{\dfrac{10}{72}+\dfrac{1}{2}}\\\\\\v_{avg}=18.78\ km/hr[/tex]

Hence , this is the required solution .

=======================================================

Work Shown:

r = radius = 1.045 cm

h = height = 30.48 cm

pi = 3.141 approximately

V = volume of cylinder

V = pi*r^2*h

V = 3.141*(1.045)^2*30.48

V = 104.547940002

V = 104.5 cubic cm

Answer:

meters, yards or feet

Explanation:

Explanation:

solution ,

[tex]a = 9.8m \div {s}^{2} [/tex]

[tex]f = 22 \: newton[/tex]

[tex] M= \: ?[/tex]

Now,

f=Ma

or,

f/a=M

or,

22/9.8=M

or,

2•24=M

or,

M=2.24kg

Answer:

e and b

Explanation:

Answer:

D,E

Explanation:

thats what I would say because PE is the energy stored in an object

Answer:

20 cm

Explanation:

Given that a ball is released from a vertical height of 20 cm. It rolls down a "perfectly frictionless" ramp and up a similar ramp. What vertical height on the second ramp will the ball reach before it starts to roll back down?

Since it is perfectly frictionless, the Kinetic energy in which the ball is rolling will be equal to the potential energy at the edge of the ramp.

Therefore, the ball will reach 20 cm before it starts to roll back down.

Answer:

Explanation:

Composite volcanoes, also called stratovolcanoes, are cone-shaped volcanoes built from many layers of lava, pumice, ash, and tephra. A lava dome is a circular mound-shaped protrusion resulting from the slow extrusion of viscous lava from a volcano. A shield volcano is a type of volcano usually composed almost entirely of fluid lava flows. It is named for its low profile, resembling a warrior's shield lying on the ground. A cinder cone is a steep conical hill of loose pyroclastic fragments, such as either volcanic clinkers, volcanic ash, or cinder that has been built around a volcanic vent.

Answer:

The correct option is;

The acceleration remains constant

Explanation:

The acceleration is due to the force of gravitational attraction between the text book and the Earth

According to Newton's law of gravitation, there is an attractive force between all objects given by the following relation;

F = G×M₁×m₂/r²

Where;

G, M₁, m₂, and r are constant such that we have;

G×M₁/r² = Constant = The acceleraton due to gravity, g

F = g×m₂

So the acceleration of the textbook as it is being attracted by the force of gravity towards the ground (Earth) is remains constant.

Answer:

7 m

Explanation:

They began walking 6 meters to the right, so +6.

Then 3 meters left so -3,

And then 4 meters to the right again so +4.

4+6-3=7 meters

Answer:

Explanation:

Using the equation of motion to get the acceleration due to gravoty of the rock on the planet.

S = ut+ 1/2at² where;

S is the distance of the rock above the surface of the planet = 100m

u is the initial velocity = 15m/s

a is the acceleration due to gravity

t is the time taken by the rock to reach the ground = 10s

Since the rock is thrown upward the acceleration due to gravity will be negative i.e a= -g

The equation becomes S = ut- 1/2gt²

Substituting the given value to get the time t

100 = 15(10)- 1/2g(10)²

100 = 150-50g

100-150 = -50g

-50 = -50g

g = -50/-50

g = 1m/s²

Hence the acceleration due to gravity of the rock when it is on Planet XX is 1m/s²

Answer: 7.27 m/s

Explanation:

Answer:

A hypothesis is a testable statement, and a theory is well-supported explanation.

Answer:

potential energy,

Explanation:

Energy, potential energy, is stored in the covalent bonds holding atoms together in the form of molecules. This is often called chemical energy.

The energy stored on the chemical bonds of molecules is chemical energy. When the chemical bonds are broken, the energy is released.

Chemical energy is a form of energy which is stored in the form of chemical bonds. There are various other forms of energy such as electrical energy, mechanical energy, kinetic energy etc.

There are different types of chemical bonds such as ionic bonds, covalent bonds, metallic bonds etc. Each type of bond are of different bond strength and energy.

Molecules are formed by the combination of atoms through chemical bonding. Chemical energy is stored within the bonds and this energy releases when the bond breaks.

To find more on chemical energy, refer here:

https://brainly.com/question/1397271

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Answer:

initial velocity is0m/s and final velocity is29.4m/s

Answer:

Where are the measurements?

Explanation:

We can measure the length of the swimming pool by using the SI unit meter, as the length of the swimming pool comes under the fundamental measurement of the dimension which is measured by the standardized international unit of length.

A unit of measurement is a specified magnitude of a quantity that is established and used as a standard for measuring other quantities of the same kind. It is determined by convention or regulation.

There are three additional units and 7 SI basic units in this system of SI units, the base SI units are meter, kilogram, second, kelvin, ampere, candela, and mole.

As given in the problem we have to find how can we measure the length of the swimming pool,

Due to the fact that the length of the swimming pool falls under the category of a dimension that is fundamentally measured using a standardized international unit of length.

Thus, we can measure the length of the swimming pool by using the SI unit meter,

Learn more about the unit of measurement from here,

brainly.com/question/12629581

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Your question seems incomplete the complete question is,

Which SI units would you use for the following measurements?

the length of a swimming pool

see this picture please mark brainlist

Answer: 1,562.5 Km

Explanation: If it takes one hour to travel 625 kilometers, then, in two and a half hours, you should travel 1,562.5 Kilometers because of 625x2.5=1562.5 kilometers

Answer:

Sound energy (that is, mechanical vibrational waves passing through an elastic medium) is a combination of kinetic and potential energies. As the wave compresses the medium (storing elastic potential energy) the medium comes to rest (no motion so no kinetic energy).

Explanation:

Hope this helps :D

Answer:

920 kg

Explanation:

[tex]kinetic \: energy = 222640j \\ \frac{1}{2} mv {}^{2} = 222640 \\ \frac{1}{2} m(22) {}^{2} = 222640 \\ m = 920kg[/tex]

Answer:

A process where you must look for terms that have identical variable parts and then combine their coefficient.

Answer:

2 electrons in energy level 6

Answer:

d. tropical

Explanation:

Tropical climates are characterized by constant high temperatures

Answer:

[tex]V_0[/tex]

Explanation:

Given that, the range covered by the sphere, [tex]M[/tex], when released by the robot from the height, [tex]H[/tex], with the horizontal speed [tex]V_0[/tex] is [tex]D[/tex] as shown in the figure.

The initial velocity in the vertical direction is [tex]0[/tex].

Let [tex]g[/tex] be the acceleration due to gravity, which always acts vertically downwards, so, it will not change the horizontal direction of the speed, i.e. [tex]V_0[/tex] will remain constant throughout the projectile motion.

So, if the time of flight is [tex]t[/tex], then

[tex]D=V_0t\; \cdots (i)[/tex]

Now, from the equation of motion

[tex]s=ut+\frac 1 2 at^2\;\cdots (ii)[/tex]

Where [tex]s[/tex] is the displacement is the direction of force, [tex]u[/tex] is the initial velocity, [tex]a[/tex] is the constant acceleration and [tex]t[/tex] is time.

Here, [tex]s= -H, u=0,[/tex] and [tex]a=-g[/tex] (negative sign is for taking the sigh convention positive in [tex]+y[/tex] direction as shown in the figure.)

So, from equation (ii),

[tex]-H=0\times t +\frac 1 2 (-g)t^2[/tex]

[tex]\Rightarrow H=\frac 1 2 gt^2[/tex]

[tex]\Rightarrow t=\sqrt {\frac {2H}{g}}\;\cdots (iii)[/tex]

Similarly, for the launched height [tex]2H[/tex], the new time of flight, [tex]t'[/tex], is

[tex]t'=\sqrt {\frac {4H}{g}}[/tex]

From equation (iii), we have

[tex]\Rightarrow t'=\sqrt 2 t\;\cdots (iv)[/tex]

Now, the spheres may be launched at speed [tex]V_0[/tex] or [tex]2V_0[/tex].

Let, the distance covered in the [tex]x-[/tex]direction be [tex]D_1[/tex] for [tex]V_0[/tex] and [tex]D_2[/tex] for [tex]2V_0[/tex], we have

[tex]D_1=V_0t'[/tex]

[tex]D_1=V_0\times \sqrt 2 t[/tex] [from equation (iv)]

[tex]\Rightarrow D_1=\sqrt 2 D[/tex] [from equation (i)]

[tex]\Rightarrow D_1=1.41 D[/tex] (approximately)

This is in the [tex]3[/tex] points range as given in the figure.

Similarly, [tex]D_2=2V_0t'[/tex]

[tex]D_2=2V_0\times \sqrt 2 t[/tex] [from equation (iv)]

[tex]\Rightarrow D_2=2\sqrt 2 D[/tex] [from equation (i)]

[tex]\Rightarrow D_2=2.82 D[/tex] (approximately)

This is out of range, so there is no point for [tex]2V_0[/tex].

Hence, students must choose the speed [tex]V_0[/tex] to launch the sphere to get the maximum number of points.

Free fall motion is motion in which the only force acting on the body is gravity

The student should launch the sphere at 2·v₀, for the sphere will land at approximately 1.41·D, which is in the 3 point zone

The given parameter are;

The distance covered by the sphere when launched at height, H = D

The velocity with which the ball reaches D = v₀

The current available height of launcher= H/2

The available velocities = v₀, and 2·v₀

Solution:

From H = u·t + (1/2)·g·t², where, initial velocity of the vertical motion of the ball, u = 0 we have;

H = (1/2)·g·t²

∴ The time it takes the ball to drop from H, t = √(2·H/g)

The distance, D = v₀ × √(2·H/g)

When the height is H/2, we get;

t = √(2·H/(2·g)) = √(H/g)

The distance covered, D₁ = v₀ × √(H/g)

Therefore, D = (√2) × v₀ × √(H/g) = (√(2))·D₁

D₁ = D/(√2) ≈ 0.71·D

D₁ ≈ 0.71·D

At speed 2·v₀, we have;

D₂ = 2·v₀ × √(H/g) = √2 × v2 × v₀ × √(H/g) = √2 × v₀ × √(2·H/g) = √2·D₁ ≈ 1.41·D

D₂ ≈ 1.41·D

The 2 point zone = D/2 < x < D = 0.5·D < x < D (Position D₁ ≈ 0.71·D is located here)

The 3 Point Zone =  D < x < 3·D/2 = D < x < 1.5·D (Position D₂ ≈ 1.41·D is located here)

Given that at D₂, the ball lands in the 3 Point Zone, the student should launch the sphere at the speed, 2·v₀, so that the ball will land at D₂ ≈ 1.41·D,  which is in the 3 Point Zone

Learn more about free fall motion here:

https://brainly.com/question/17842118

Explanation:

Answer:

A. The paper airplane’s motion is due to horizontal inertia and the vertical pull of gravity.

Explanation:

Got it right on edge.