All categories

2 years ago

Find acceleration. Will give brainliest!

Physics

2 answers:

Art [367]2 years ago

8 0

Answer:

$\boxed{\mathrm{4.5 \: kmph/s \: or \: 1.25 \: m/s^2 }}$

Explanation:

$\displaystyle \mathrm{acceleration = \frac{change \: in \: velocity}{time \: taken}}$

$\displaystyle \mathrm{a = \frac{v - u}{t}}$

$\displaystyle \mathrm{v=final \: velocity}\\\displaystyle \mathrm{u=initial \: velocity}$

$\displaystyle \mathrm{a = \frac{45- 0}{10}}$

$\displaystyle \mathrm{a = \frac{45}{10}}$

$\displaystyle \mathrm{a = 4.5}$

$\mathrm{4.5 \: kmph/s = 1.25 \: m/s^2 }$

zalisa [80]2 years ago

7 0

Answer:

16200 km/s

270 km/min

4.5 km/h

Explanation:

Acceleration Formula: Average Acceleration = Δv/Δt (change in velocity over change in time)

Simply plug in our known variables and solve:

a = (45.0 - 0)/10

a = 45.0/10

a = 4.5 km/h

You might be interested in

A tiny 0.0250 -microgram oil drop containing 15 excess electrons is suspended between to horizontally closely-spaced metal plate

qaws [65]

Answer:

(a) 12 × 10⁻³ C = 12 mC (b) The lower plate

Explanation:

Given

mass of oil drop, m = 0.025 μg = 0.025 × 10⁻⁶

radius of plates, r = 6.50 cm = 6.5 × 10⁻² m

k = 1/4πε₀ = 9.0 × 10⁹ Nm²/C²

electric charge, e= 1.6 × 10⁻¹⁹ C

charge on oil drop, q = 15e

charge on plates, Q = ?

First, we find the charge density of the plates, D = Q/A where Q = charge on plates and A = area of plates. Since the plates are circular, the area is given by A=πr² where r = radius of plates. D=Q/πr²

Also, the electric field, E between the plates is given by E = D/ε =Q/Aε = Q/ε₀πr².

The force on the oil drop due to the electric field between the plates is given by F = qE = qQ/ε₀πr².

Since the oil drop is suspended between the plates, it means that the electric force due to the field on the oil drop balances the weight of the oil drop. So, since weight of oil drop W = mg where g = 9.8 m/s². F =W (for oil drop suspension).

So, qQ/ε₀πr²=mg

So, Q=mgε₀πr²/q

From k = 1/4πε₀, ε₀=1/4πk

So, Q = mgπr²/4πkq = mgr²/4kq = (0.025 × 10⁻⁶ × 9.8 × (6.5 × 10⁻²)²)÷(4 × 9 × 10⁹ × 15 × 1.6 × 10⁻¹⁹)= 0.012 C = 12 mC

(b) The lower plate must be positive because, the direction of the electric field must be upwards, so as to balance out the weight of the oil drop so as to suspend it.

6 0

3 years ago

Which region of a longitudinal wave is this?

ipn [44]

I think the correct answer is is D.

8 0

2 years ago

Bernoulli's principle can be used to explain the lift force on an airplane wing. How must an airplane's wing be designed to ensu

Gnesinka [82]

Answer:

D. Airplane wings must be designed to ensure that air molecules move more rapidly over the top surface of the wing, creating a region of lower pressure.

Explanation:

Bernoulli's theory operates on the assumption that its shape makes the air travel more over the top of the wing than under it as a plane moves through the air. Due to the pressure difference, wings feel a lifting force and create lift force. when pressure decrease than the velocity of the fluid will increase.

Therefore the answer is D.

6 0

3 years ago

What mass of electrons would be required to just neutralize the charge of 4.8 g of protons?

suter [353]

Let's calculate the total charge of M=4.8 g=0.0048 kg of protons.
Each proton has a charge of $q=1.6 \cdot 10^{-19} C$ , and a mass of $m_p = 1.67 \cdot 10^{-27}kg$ . So, the number of protons is
$N_p = \frac{M}{m_p}= \frac{0.0048 kg}{1.67 \cdot 10^{-27}kg}=2.87 \cdot 10^{24}$ And so the total charge of these protons is $Q_p = qN_p = (1.6 \cdot 10^{-19}C)(2.87 \cdot 10^{24})=4.6\cdot 10^5 C$

So, the neutralize this charge, we must have $N_e$ electrons such that their total charge is
$Q_e = -4.6 \cdot 10^5 C$
Since the charge of each electron is $q_e = -1.6 \cdot 10^{-19}C$ , the number of electrons needed is
$N_e = \frac{Q_e}{q}= \frac{-4.6 \cdot 10^5 C}{-1.6 \cdot 10^{-19}C}=2.87 \cdot 10^{24}$
which is the same as the number of protons (because proton and electron have same charge magnitude). Since the mass of a single electron is $m_e=9.1 \cdot 10^{-31}kg$ , the total mass of electrons should be
$M_e = N_e m_e = (2.87 \cdot 10^{24})(9.1 \cdot 10^{-31}kg)=2.6 \cdot 10^{-6}kg$

6 0

2 years ago

(a) Triply charged uranium-235 and uranium-238 ions are being separated in a mass spectrometer. (The much rarer uranium-235 is u

stiv31 [10]

Answer:

(a) 2.5 cm

(b) Yes

Solution:

As per the question:

Mass of Uranium-235 ion, m = $3.95\times 10^{- 25}\ kg$

Mass of Uranium- 238, m' = $3.90\times 10^{- 25}\ kg$

Velocity, v = $3.00\times 10^{5}\ m/s$

Magnetic field, B = 0.250 T

q = 3e

Now,

To calculate the path separation while traversing a semi-circle:

$\Delta x = 2(R_{U_{35}} - 2R_{U_{38}})$

The radius of the ion in a magnetic field is given by:

R = $\frac{mv}{qB}$

$\Delta x = 2(R_{U_{35}} - 2R_{U_{38}})$

$\Delta x = 2(\frac{mv}{qB} - \frac{m'v}{qB})$

$\Delta x = 2(\frac{m - m'}{qB}v)$

Now,

By putting suitable values in the above eqn:

$\Delta x = 2(\frac{3.95\times 10^{- 25} - 3.90\times 10^{- 25}}{3\times 1.6\times 10^{- 19}\times 0.250}\times 3.00\times 10^{5}) = 2.5\ cm$

$\Delta x = 1.25\ cm$

(b) Since the order of the distance is in cm, thus clearly this distance is sufficiently large enough in practical for the separation of the two uranium isotopes.

3 0

2 years ago