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djverab [1.8K]
2 years ago
5

A crate is sliding down an inclined ramp at a constant speed of 0.55 m/s. The vector sum of all the forces acting on this crate

must point down the ramp
Physics
1 answer:
Lemur [1.5K]2 years ago
8 0

Answer:

F_{net} = 0

Explanation:

here it is given that the crate is sliding down with constant speed

so we will have

v = constant

now we can say

a = \frac{dv}{dt}

so here we have

a = 0

now from Newton's II law we can say that sum of all forces on the system is always product of mass and its acceleration

so here we will have

F_{net} = ma

F_{net} = 0

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A football player kicks a football downfield. The height of the football increases until it reaches a maximum height of 15 yards
Trava [24]

Answer:

kick 1 has travelled 15 + 15 = 30 yards before hitting the ground

so kick 2 travels 25 + 25 = 50 yards before hitting the ground

first kick reached 8 yards and 2nd kick reached 20 yards  

Explanation:

1st kick travelled 15 yards to reach maximum height of 8 yards

so, it has travelled 15 + 15 = 30 yards before hitting the ground

2nd kick is given by the equation

y (x) = -0.032x(x - 50)

Y = 1.6 X - 0.032x^2

we know that maximum height occurs is given as

x = -\frac{b}{2a}

y =- \frac{1.6}{2(-0.032)} = 25

and maximum height is

y = 1.6\times 25 - 0.032\times 25^2

y = 20

so kick 2 travels 25 + 25 = 50 yards before hitting the ground

first kick reached 8 yards and 2nd kick reached 20 yards

8 0
2 years ago
Read 2 more answers
A toy rocket engine is securely fastened to a large puck that can glide with negligible friction over a horizontal surface, take
Ad libitum [116K]

Answer:

F=(3i+3.6j)\ N

Explanation:

It is given that,

Mass of the puck, m = 4.8 kg

Initial velocity of the puck, u=(1i+0j)\ m/s

After 8 seconds, final velocity of the puck, v=(6i+6j)\ m/s

Let the x and y component of force is given by F_x\ and\ F_y.

x component of force is given by :

F_x=m\times \dfrac{v-u}{t}

F_x=4.8\times \dfrac{6-1}{8}

F_x=3\ N

y component of force is given by :

F_y=m\times \dfrac{v-u}{t}

F_y=4.8\times \dfrac{6-0}{8}

F_y=3.6\ N

So, the component of the force is F=(3i+3.6j)\ N. Hence, this is the required solution.

7 0
2 years ago
At what height h above the ground does the projectile have a speed of 0.5v?
maw [93]

Answer:

h=\dfrac{3v^2}{8g}

Explanation:

It is given that,

Speed of the projectile is 0.5 v. Let h is the height above the ground. Using the first equation of motion to find it.

v=u+at

v=u-gt

Initial speed of the projectile is v and final speed is 0.5 v.

0.5v=v-gt

t=\dfrac{v}{2g}

g is the acceleration due to gravity

Let h is the height above the ground. Using the second equation of motion as :

h=vt-\dfrac{1}{2}gt^2

h=v\dfrac{v}{2g}-\dfrac{1}{2}g(\dfrac{v}{2g})^2

h=\dfrac{3v^2}{8g}

So, the height of the projectile above the ground is \dfrac{3v^2}{8g}. Hence, this is the required solution.

6 0
1 year ago
A sample of 4.50 g of methane occupies 12.7 dm3 at 310 K. (a) Calculate the work done when the gas expands isothermally against
valkas [14]

Answer:

(A) Work done will be 87.992 KJ

(B) Work done will be 167.4 KJ            

Explanation:

We have given mass of methane m = 4.5 gram = 0.0045 kg

Volume occupies V_1=12.7dm^3=12.7liters

And volume is increased by 3.3dm^3 so V_2=12.7+3.3=16liters

Temperature T = 310 K

Pressure is given as 200 Torr = 26664.5 Pa

(a) At constant pressure work done is given by

W=P(V_2-V_1)=26664.5\times (16-12.7)=87992.85J=87.992kj

(b) At reversible process work done is given by W=nRTln\frac{V_2}{V_1}

We have given mass = 4.5 gram

Molar mass of methane = 16

So number of moles n=\frac{mass\ in\ gram}{mol;ar\ mass}=\frac{4.5}{16}=0.28125

So work done W=0.28125\times 8.314\times 310ln\frac{16}{12.7}=167.4J

7 0
1 year ago
Assuming no air resistance, how far will a 0.0010 kg raindrop have fallen when it hits the ground 30.0 s later. Assume g = 9.8 m
Serga [27]
D = (1/2)·at²
where d is the distance fallen, a is the acceleration (g in this problem), and t is the time
d = (1/2)·(9.8 m/s²)·(30 s)² = (1/2)·(9.8)·(900) m
d = 4410 m
The answer is b) 4410 m

Note: the mass of the raindrop is irrelevant since the acceleration due to gravity is independent of mass. (Galileo's Leaning Tower of Pisa experiment)
6 0
2 years ago
Read 2 more answers
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