A car acquires a velocity of 72km/h in 10sec starting from rest .find it's

Which of these is not true about the proton?

bulgar [2K]

Answer:

Option C is the untrue statement.

5 0

4 years ago

Read 2 more answers

If a ball is thrown vertically upward with a velocity of 80 ft/s, then its height after t seconds is s = 80t − 16t 2 . (a) what

MAXImum [283]

Max height occurs when v = 0.
v(t) = ds(t)/dt
v(t) = 80 - 32t
0 = 80 - 32t
t = 5/2

s(5/2) = 80(5/2) - 16(5/2)^2
s(5/2) = 100
Answer: 100 ft

96 = 80t - 16t²
t = 3, 2
(80 ± √256) / 32 using the quadratic equation.

v(2) = 16
v(3) = -16

4 0

3 years ago

A 06-C charge and a .07-C charge are apart at 3 m apart. What force attracts them?

Andru [333]

Answer:

F = 37.8 × 10^(6) N

Explanation:

The charges are 0.06 C and 0.07 C.

Thus;

Charge 1; q1 = 0.06 C

Charge 2; q2 = 0.07 C

Distance between them; r = 3 m

Formula for the force in between them is;

F = kq1•q2/r²

Where k is a constant = 9 × 10^(9) N.m²/C²

Thus;

F = (9 × 10^(9) × 0.06 × 0.07)/3²

F = 37.8 × 10^(6) N

3 0

3 years ago

Nuclear

luda_lava [24]

Answer:

Energy

Kinetic

Energy in

this

Explanation:

ithikitsthatecauseireallydo

4 0

3 years ago

A cylinder with a piston contains 0.200 mol of oxygen at 2.00×105 Pa and 360 K . The oxygen may be treated as an ideal gas. The

ikadub [295]

Answer:

A. $W=600\ J$

B. $Q=2112\ J$

C. $\Delta U=1512\ J$

D. $W=0\ J$

Explanation:

Given:

no. of moles of oxygen in the cylinder, $n=0.2$

initial pressure in the cylinder, $P_i=2\times 10^5\ Pa$

initial temperature of the gas in the cylinder, $T_i=360\ K$

According to the question the final volume becomes twice of the initial volume.

Using ideal gas law:

$P.V=n.R.T$

$2\times 10^5\times V_i=0.2\times 8.314\times 360$

$V_i=0.003\ m^3$

A.

Work done by the gas during the initial isobaric expansion:

$W=P.dV$

$W=P_i\times (V_f-V_i)$

$W=2\times 10^5\times (0.006-0.003)$

$W=600\ J$

C.

we have the specific heat capacity of oxygen at constant pressure as:

$c_v=21\ J.mol^{-1}.K^{-1}$

Now we apply Charles Law:

$\frac{V_i}{T_i} =\frac{V_f}{T_f}$

$\frac{0.003}{360} =\frac{0.006}{T_f}$

$T_f=720\ K$

Now change in internal energy:

$\Delta U=n.c_p.(T_f-T_i)$

$\Delta U=0.2\times 21\times (720-360)$

$\Delta U=1512\ J$

B.

Now heat added to the system:

$Q=W+\Delta U$

$Q=600+1512$

$Q=2112\ J$

D.

Since during final cooling the process is isochoric (i.e. the volume does not changes). So,

$W=0\ J$

7 0

3 years ago