Throw a paper ball towards a hard wall.

On a perfect fall day, you are hovering at low altitude in a hot-air balloon, accelerated neither upward nor downward. The total

Stella [2.4K]

Explanation:

Since the balloon is not accelerating means that the net force on the balloon is zero. This implies that the weight of balloon must be equal to the buoyant force on balloon.

Hence, the buoyant force equals the weight of air displaced by the balloon, also 20,000 N.

Weight of the air displaced = density of air × volume

The density of air at 1 atm pressure and 20º C is 1.2 kg/m³

the volume V = 20,000/(1.2×9.8) = 1700 m³

3 0

3 years ago

How much heat is necessary to warm 500g of water from 20°c to 65°c

Vsevolod [243]

Explanation:

so sorry

don't know but please mark me as brainliest please

8 0

3 years ago

g A 1.5-kg mass attached to spring with a force constant of 20.0 N/m oscillates on a horizontal, frictionless track. At t = 0, t

jok3333 [9.3K]

Answer:

(a) f = 0.58Hz

(b) vmax = 0.364m/s

(c) amax = 1.32m/s^2

(d) E = 0.1J

(e) x(t)=0.1m*cos(2π(0.58s^{-1})t)

Explanation:

(a) The frequency of the oscillation, in a spring-mass system, is calculated by using the following formula:

$f=\frac{1}{2\pi}\sqrt{\frac{k}{m}}$ (1)

k: spring constant = 20.0N/m

m: mass = 1.5kg

you replace the values of m and k for getting f:

$f=\frac{1}{2\pi}\sqrt{\frac{20.0N/m}{1.5kg}}=0.58s^{-1}=0.58Hz$

The frequency of the oscillation is 0.58Hz

(b) The maximum speed is given by the following relation:

$v_{max}=\omega A=2\pi f A$ (2)

A: amplitude of the oscillations = 10.0cm = 0.10m

$v_{max}=2\pi (0.58s^{-1})(0.10m)=0.364\frac{m}{s}$

The maximum speed of the mass is 0.364 m/s.

The maximum speed occurs when the mass passes trough the equilibrium point of the oscillation.

(c) The maximum acceleration is given by the following formula:

$a_{max}=\omega^2A=(2\pi f)^2 A$

$a_{max}=(2\pi (0.58s^{-1}))(0.10m)=1.32\frac{m}{s^2}$

The maximum acceleration is 1.32 m/s^2

The maximum acceleration occurs where the elastic force is a maximum, that is, where the mass is at the maximum distance from the equilibrium point, that is, the acceleration.

(d) The total energy of the system is calculated with the maximum potential elastic energy:

$E=\frac{1}{2}kA^2=\frac{1}{2}(20.0N/m)(0.10m)^2=0.1J$

The total energy is 0.1J

(e) The displacement as a function of time is:

$x(t)=Acos(\omega t)=Acos(2\pi ft)\\\\x(t)=0.1m\ cos(2\pi(0.58s^{-1})t)$

6 0

3 years ago

A bullet with a mass of 5 gramsand speed of 560 m/sis fired horizontally atablock of wood with a mass of 2 kg. The block rests o

Anni [7]

Momentum is conserved if and only if sum of all forces which are exserted on system equals zero. In our situation there are only internal forces, so by Newton's third law their vector sum is 0.

So $mv=(m+M)v' \Leftrightarrow v'=\frac{mv}{m+M}$ .

Kinetic energy of system at first: $\frac{mv^2}{2}=784\;\textbf{J}$ . After: $\frac{(m+M)\frac{m^2v^2}{(m+M)^2} }{2}=\frac{m^2v^2}{2(m+M)}\approx 1,96\; \textbf{J}$ . The secret is that other energy is in work of deformation forces (they in turn heat a bullet and a block).

Answer is A)

5 0

3 years ago

A worker pushes horizontally on a large crate with a force of 200 N, and the crate is moved 3.5 m. How much work was done in Jou

sattari [20]

Work = (force) x (distance) =

(200 N) x (3.5 m) = <em>700 joules</em>

6 0

3 years ago