How much does a 60 kg table weigh

A cannon fires a cannonball 500.0 m downrange when set at a 45.0o angle. At what velocity does the cannonball leave the cannon?

kobusy [5.1K]

Answer:

v = 70 m/s

Explanation:

Range on the cannon ball is given as

$d = 500.0 m$

here the angle of the projection of the ball is given as 45 degree

now we know that if the velocity of the ball is "v" then its two components will be given as

$v_x = vcos45$

$v_y = vsin45$

so here time of flight of the motion is given as

$T = \frac{2v_y}{g}$

$T = \frac{2vsin45}{g}$

also the range is given as

$R = v_x T$

$R = (vcos45)(\frac{2vsin45}{g})$

now plug in all data in this equation

$500.0 = \frac{v^2(2sin45cos45)}{g}$

$v = 70 m/s$

4 0

3 years ago

An object of height 2.7 cm is placed 29 cm in front of a diverging lens of focal length 16 cm. Behind the diverging lens, and 12

kati45 [8]

Answer:

A) q = -8.488 cm , B) m = 0.29

Explanation:

A) For this exercise in geometric optics, we will use the equation of the constructor

$\frac{1}{f} = \frac{1}{p} + \frac{1}{q}$

where p and q are the distance to the object and image, respectively and f is the focal length

in our case the distance the object is p = 29 cm the focal length of a diverging lens is negative and indicates that it is f = - 12 cm

$\frac{1}{q} = \frac{1}{f} - \frac{1}{p}$

we calculate

$\frac{1}{q} = - \frac{1}{12} - \frac{1}{29}$

$\frac{1}{q}$ = - 0.1178

q = -8.488 cm

the negative sign indicates that the image is virtual

B) the magnification is given

$m = \frac{h'}{h} = - \frac{q}{p}$

we substitute

m = $- \frac{-8.488}{29}$

m = 0.29

the positive sign indicates that the image is right

4 0

3 years ago

Evangelista Torricelli was the first person to realize that we live at the bottom of an ocean of air. He correctly surmised that

Radda [10]

Answer:

a) h = 8.02 10³ m b) yes

Explanation:

a) The pressure in a fluid is given by

P = ρ g h

The pressure in this case is the atmospheric pressure, 1.013 105 Pa, let's clear the height (h)

h = P / ρ g

h = 1.013 10⁵ / (1.29 9.8)

h = 8.02 10³ m

b) The height of Mount Everest is 8848 m

It is above this height, according to this model there would be no air to breathe

8 0

2 years ago

A coil with an inductance of 2.3 H and a resistance of 14 Ω is suddenly connected to an ideal battery with ε = 100 V. At 0.13 s

klemol [59]

Given Information:

Resistance = R = 14 Ω

Inductance = L = 2.3 H

voltage = V = 100 V

time = t = 0.13 s

Required Information:

(a) energy is being stored in the magnetic field

(b) thermal energy is appearing in the resistance

(c) energy is being delivered by the battery?

Answer:

(a) energy is being stored in the magnetic field ≈ 219 watts

(b) thermal energy is appearing in the resistance ≈ 267 watts

(c) energy is being delivered by the battery ≈ 481 watts

Explanation:

The energy stored in the inductor is given by

$U = \frac{1}{2} Li^{2}$

The rate at which the energy is being stored in the inductor is given by

$\frac{dU}{dt} = Li\frac{di}{dt} \: \: \: \: eq. 1$

The current through the RL circuit is given by

$i = \frac{V}{R} (1-e^{-\frac{t}{ \tau} })$

Where τ is the the time constant and is given by

$\tau = \frac{L}{R}\\ \tau = \frac{2.3}{14}\\ \tau = 0.16$

$i = \frac{110}{14} (1-e^{-\frac{t}{ 0.16} })\\i = 7.86(1-e^{-6.25t})\\\frac{di}{dt} = 49.125e^{-6.25t}$

Therefore, eq. 1 becomes

$\frac{dU}{dt} = (2.3)(7.86(1-e^{-6.25t}))(49.125e^{-6.25t})$

At t = 0.13 seconds

$\frac{dU}{dt} = (2.3) (4.37) (21.8)\\\frac{dU}{dt} = 219.11 \: watts$

(b) thermal energy is appearing in the resistance

The thermal energy is given by

$P = i^{2}R\\P = (7.86(1-e^{-6.25t}))^{2} \cdot 14\\P = (4.37)^{2}\cdot 14\\P = 267.35 \: watts$

(c) energy is being delivered by the battery?

The energy delivered by battery is

$P = Vi\\P = 110\cdot 4.37\\P = 481 \: watts$

4 0

3 years ago

What evidence supports the conclusion that a chemical reaction has occurred?

GalinKa [24]

Answer:

Precipitation formed

Temperature change

fizzing and hissing

bubbles

change in state

Explanation:

3 0

2 years ago