All categories

3 years ago

Which equation is most likely used to determine the acceleration from a velocity vs:time graph?

2 answers:

7 0

Acceleration, a = (v - u)/t

where v is the final velocity, u is the initial velocity, and t is the time.

This formula on a velocity time graph represents the slope of the graph.

Mars2501 [29]3 years ago

3 0

Answer: Hello there!

in this case, you have a graph of velocity vs time, then in this graph is represented the function v(t).

The acceleration is defined as the integral with respect to the time of the velocity v(t), then we could calculate the mean acceleration between two times in a next way:

$a = \frac{v(t2) - v(t1))}{t2 -t1}$

this is the slope between the times t2 and t1, where t2>t1 and V(t2) is the velocity at the time t2 (the final velocity) and v(t1) is the velocity at the time t1 (the initial velocity). In this case, a is the mean acceleration between these times.

You might be interested in

The anther and filament are parts of a flower's:

zaharov [31]

A the stamen :)). haha yea smert

6 0

3 years ago

Read 2 more answers

It is known that the gravitational force of attraction between two alpha particles is much weaker than the electrical repulsion.

natali 33 [55]

Answer:

The ratio of gravitational force to electrical force is 3.19 x 10^-36

Explanation:

mass of an alpha particle = 6.64 x $10^{-27}$ kg

charge on an alpha particle = +2e = +2(1.6 x $10^{-19}$ C) = 3.2 x $10^{-19}$ C

distance between particles = d

For gravitational attraction:

The force of gravitational attraction F = $\frac{Gm^{2} }{r^{2} }$

where G = gravitational constant = 6.67 x $10^{-11}$ m^3 kg^-1 s^-2

r = the distance between the particles = d

m = the mass of each particle

therefore, gravitational force = $\frac{6.67*10^{-11}*(6.64*10^{-27} )^{2} }{d^{2} }$ = $\frac{2.94*10^{-63} }{d^{2} }$ Newton

For electrical repulsion:

Electrical force between the particles = $\frac{-kQ^{2} }{r^{2} }$

where k is the Coulomb's constant = 9.0 x $10^{9}$ N•m^2/C^2

r = distance between the particles = d

Q = charge on each particle

therefore, electrical force = $\frac{-9*10^{9}*(3.2*10^{-19} )^{2} }{d^{2} }$ = $\frac{-9.216*10^{-28} }{d^{2} }$ Newton

the negative sign implies that there is a repulsion on the particles due to their like charges.

Ratio of the magnitude of gravitation to electrical force = $\frac{2.94*10^{-63} }{9.216*10^{-28} }$

==> 3.19 x 10^-36

8 0

3 years ago

What are the density, specific gravity and mass of the air in a room whose dimensions are 4 m * 6 m * 8 m at 100 kPa and 25 C.

Volgvan

Answer:

Density = 1.1839 kg/m³

Mass = 227.3088 kg

Specific Gravity = 0.00118746 kg/m³

Explanation:

Room dimensions are 4 m, 6 m & 8 m. Thus, volume = 4 × 6 × 8 = 192 m³

Now, from tables, density of air at 25°C is 1.1839 kg/m³

Now formula for density is;

ρ = mass(m)/volume(v)

Plugging in the relevant values to give;

1.1839 = m/192

m = 227.3088 kg

Formula for specific gravity of air is;

S.G_air = density of air/density of water

From tables, density of water at 25°C is 997 kg/m³

S.G_air = 1.1839/997 = 0.00118746 kg/m³

4 0

3 years ago

A dragster race car can accelerate from rest to incredible speeds. In one case a dragster is able to finish the 305 m run in 3.6

Dafna1 [17]

Answer:

45.89m/s²

Explanation:

Given

Distance S = 305m

Time t = 3.64s

To get the acceleration during this run, we will apply the equation of motion:

S = ut+1/2at²

Substitute the given parameters into the formula and calculate the value of a

305 = 0+1/2 a(3.64)²

304 = 1/2(13.2496)a

304 = 6.6248a

a = 304/6.6248

a = 45.89m/s²

Hence the average acceleration during this run is 45.89m/s²

4 0

3 years ago

In an elastic head-on collision, a 0.60 kg cart moving at 5.0 m/s [W] collides with a 0.80 kg cart moving at 2.0 m/s [E]. The co

labwork [276]

Answer:

The answer is given below

Explanation:

u is the initial velocity, v is the final velocity. Given that:

$m_1=0.6kg,u_1=-5m/s(moving \ west),m_2=0.8kg,u_2=2m/s,k=1200N/m$

The final velocity of cart 1 after collision is given as:

$v_1=(\frac{m_1-m_2}{m_1+m_2})u_1+\frac{2m_2}{m_1+m_2}u_2\\ Substituting:\\v_1=\frac{0.6-0.8}{0.6+0.8} (-5)+\frac{2*0.8}{0.6+0.8}(2)= 5/7+16/7=3\ m/s$

The final velocity of cart 2 after collision is given as:

$v_2=(\frac{m_2-m_1}{m_1+m_2})u_2+\frac{2m_1}{m_1+m_2}u_1\\ Substituting:\\v_1=\frac{0.8-0.6}{0.6+0.8} (2)+\frac{2*0.6}{0.6+0.8}(-5)= 2/7-30/7=-4\ m/s$

b) Using the law of conservation of energy:

$\frac{1}{2}m_1u_1+ \frac{1}{2}m_2u_2=\frac{1}{2}m_1v_1+\frac{1}{2}m_2v_2+\frac{1}{2}kx^2\\x=\sqrt{\frac{m_1u_1+m_2u_2-m_1v_1-m_2v_2}{k}}\\ Substituting\ gives:\\x=\sqrt{\frac{0.6*(-5)^2+0.8*2^2-(0.6*3^2)-(0.8*(-4)^2)}{1200}}=\sqrt{0}=0\ cm$

7 0

3 years ago