VERY EASY QUESTION FOR HIGH SCHOOL STUDENTS:

Will is a scientist. He’s designing a spacecraft that would allow people to land on Mars. Will’s mass on Earth is 75 kilograms.

Vesnalui [34]

Mass will remain unchanged, always. His weight, which is the gravitational force acting on that mass will be less in this case.

6 0

3 years ago

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A jogger runs 6 km north ,5 km east then another 4km north her average speed 8 km how long will it take her to complete her run

LekaFEV [45]

The time taken to complete her run is 1.9 hr.

<u>Explanation:</u>

Speed is a scalar quantity and it is defined as the ratio of distance covered to the time taken to cover that distance. As distance is also a scalar quantity, so the directions given in the problem can be ignored. Thus, the distance covered by the jogger is the sum of kilometers given in problem.

Distance covered = 6+5+4 = 15 km

And the speed is given as 8 km/hr.

So the time taken will be ratio of distance to speed.

$\text { time }=\frac{\text {distance}}{\text {speed}}=\frac{15}{8}=1.9 \text { hour }$

So the jogger will take 1.9 hr to complete her run.

8 0

4 years ago

According to the Big Bang theory how long ago did the universe started??

Rashid [163]

According to the Big Bang theory how long ago did the universe started??
13 to 15 billion years ago

7 0

3 years ago

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A projectile is fired into the air from the top of a 200-m cliff above a valley as shown below. Its initial velocity is 60 m/s a

anastassius [24]

a) y(max) = 337.76 m

b) t₁ = 5.30 s the time for y maximum

c)t₂ = 13.60 s time for y = 0 time when the fly finish

d) vₓ = 30 m/s vy = - 81.32 m/s

e)x = 408 m

Equations for projectile motion:

v₀ₓ = v₀ * cosα v₀ₓ = 60*(1/2) v₀ₓ = 30 m/s ( constant )

v₀y = v₀ * sinα v₀y = 60*(√3/2) v₀y = 30*√3 m/s

a) Maximum height:

The following equation describes the motion in y coordinates

y = y₀ + v₀y*t - (1/2)*g*t² (1)

To find h(max), we need to calculate t₁ ( time for h maximum)

we take derivative on both sides of the equation

dy/dt = v₀y - g*t

dy/dt = 0 v₀y - g*t₁ = 0 t₁ = v₀y/g

v₀y = 60*sin60° = 60*√3/2 = 30*√3

g = 9.8 m/s²

t₁ = 5.30 s the time for y maximum

And y maximum is obtained from the substitution of t₁ in equation (1)

y (max) = 200 + 30*√3 * (5.30) - (1/2)*9.8*(5.3)²

y (max) = 200 + 275.40 - 137.64

y(max) = 337.76 m

Total time of flying (t₂) is when coordinate y = 0

y = 0 = y₀ + v₀y*t₂ - (1/2)* g*t₂²

0 = 200 + 30*√3*t₂ - 4.9*t₂² 4.9 t₂² - 51.96*t₂ - 200 = 0

The above equation is a second-degree equation, solving for t₂

t = [51.96 ±√ (51.96)² + 4*4.9*200]/9.8

t = [51.96 ±√2700 + 3920]/9.8

t = [51.96 ± 81.36]/9.8

t = 51.96 - 81.36)/9.8 we dismiss this solution ( negative time)

t₂ = 13.60 s time for y = 0 time when the fly finish

The components of the velocity just before striking the ground are:

vₓ = v₀ *cos60° vₓ = 30 m/s as we said before v₀ₓ is constant

vy = v₀y - g *t vy = 30*√3 - 9.8 * (13.60)

vy = 51.96 - 133.28 vy = - 81.32 m/s

The sign minus means that vy change direction

Finally the horizontal distance is:

x = vₓ * t

x = 30 * 13.60 m

x = 408 m

5 0

3 years ago

A grinding wheel is a uniform cylinder with a radius of 8.65 cm and a mass of 0.400 kg . Calculate its moment of inertia about i

anyanavicka [17]

Answer:

I = 1.5*10⁻³ kg*m²

Explanation:

It can be showed that the moment of inertia (or rotational inertia) for a uniform cylinder of mass m and radius r, respect an longitudinal axis going through its center (parallel to the height of the cylinder) can be written as follows:

$I = \frac{1}{2}*m*r^{2} = \frac{1}{2}*0.400 kg*(0.0865m)^{2} = 1.5e-3 kg*m2$

3 0

3 years ago