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3 years ago

Number these from least (1) to most (5) mass ?

Physics

2 answers:

elena-14-01-66 [18.8K]3 years ago

6 0

1. a feather

2. a baseball

3.a small car

4. a truck

5 a large train

alisha [4.7K]3 years ago

5 0

Answer:

1. A feather

2. A baseball

3. A small car

4. A truck

5. A large train

Explanation:

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For many years Colonel John P. Stapp, USAF, held the world's land speed record. He participated in studying whether a jet pilot

Margarita [4]

Answer:

(a) -202 m/s²

(b) 198 m

Explanation:

Given data

Initial speed (v₀): 283 m/s

$\frac{632mi}{h} .\frac{1609.34m}{1mi} .\frac{1h}{3600s} =283m/s$

Final speed (vf): 0 (rest)

Time (t): 1.40 s

(a) The acceleration (a) is the change in the speed over the time elapsed.

a = (vf - v₀)/t = (0 - 283 m/s)/ 1.40s = -202 m/s²

(b) We can find the distance traveled (d) using the following kinematic expression.

y = v₀ × t + 1/2 × a × t²

y = 283 m/s × 1.40 s + 1/2 × (-202 m/s²) × (1.40 s)²

y = 198 m

3 0

3 years ago

Q1 is located at the origin, Q2 is located at x = 2.50 cm and Q3 is located at x = 3.50 cm. Q1 has a charge of +4.92μC and Q3 ha

Inessa05 [86]

Answer:

$+1.11\mu C$

Explanation:

A charge located at a point will experience a zero electrostatic force if the resultant electric field on it due to any other charge(s) is zero.

$Q_1$ is located at the origin. The net force on it will only be zero if the resultant electric field intensity due to $Q_2$ and $Q_3$ at the origin is equal to zero. Therefore we can perform this solution without necessarily needing the value of $Q_1$ .

Let the electric field intensity due to $Q_2$ be + $E_2$ and that due to $Q_3$ be - $E_3$ since the charge is negative. Hence at the origin;

$+E_2-E_3=0..................(1)$

From equation (1) above, we obtain the following;

$E_2=E_3.................(2)$

From Coulomb's law the following relationship holds;

$+E_2=\frac{kQ_2}{r_2^2}\\$

$-E_3=\frac{kQ_3}{r_3^2}$

where $r_2$ is the distance of $Q_2$ from the origin, $r_3$ is the distance of $Q_3$ from the origin and k is the electrostatic constant.

It therefore means that from equation (2) we can write the following;

$\frac{kQ_2}{r_2^2}=\frac{kQ_3}{r_3^2}.................(3)$

k can cancel out from both side of equation (3), so that we finally obtain the following;

$\frac{Q_2}{r_2^2}=\frac{Q_3}{r_3^2}................(4)$

Given;

$Q_2=?\\r_2=2.5cm=0.025m\\Q_3=-2.18\mu C=-2.18* 10^{-6}C\\r_3=3.5cm=0.035m$

Substituting these values into equation (4); we obtain the following;

$\frac{Q_2}{0.025^2}=\frac{2.18*10^{-6}}{0.035^2}\\\\hence;\\\\Q_2=\frac{0.025^2*2.18*10^{-6}}{0.035^2}\\$

$Q_2=\frac{0.00136*10^{-6}}{0.00123}=1.11*10^{-6}C\\\\Q_3=+1.11\mu C$

6 0

3 years ago

Find the ratio of the final speed of the electron to the final speed of the hydrogen ion, assuming non-relativistic speeds. Take

KiRa [710]

Answer:

$\frac{V_{e}}{V_{h}}=0.428*10^{2}$

Explanation:

From conservation of energy states that

$K_{i}+v_{i}=v_{f}+K_{f}\\ as\\K_{i}=0\\K_{f}=1/2mv^{2}\\ v_{i}=qv\\v_{f}=0\\So\\qv=1/2mv^{2}\\ v=\sqrt{\frac{2qv}{m} }\\ Velocity_{electron}=\sqrt{\frac{2qv}{m_{e}} }\\Velocity_{hydrogen}=\sqrt{\frac{2qv}{m_{h}} }\\\frac{V_{e}}{V_{h}}=\sqrt{\frac{\frac{2qv}{m_{e}}}{\frac{2qv}{m_{h}}}}\\\frac{V_{e}}{V_{h}}=\sqrt{\frac{m_{h}}{m_{e}} }\\\frac{V_{e}}{V_{h}}=\sqrt{\frac{1.67*10^{-27} }{9.11*10^{-31} } }\\\frac{V_{e}}{V_{h}}=0.428*10^{2}$

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3 years ago

An elevator accelerates upward at 1.2 m/s 2 . the acceleration of gravity is 9.8 m/s 2 . what is the upward force exerted by the

Marysya12 [62]

We can find the force by using the following formula;
N = ma + mg
Fa = ma = 76 x 1.2 = 91.2
Fg = mg = 76 x 9.8 = 744.8
N = 91.2 + 744.8 = 836
So, the force is 836 N.

8 0

3 years ago

a concrete slab 20 m long and weighing 400,000 N is supported by one pillar. A 19,600 N car is parked 8 meters from one end, whe

Elden [556K]

let the distance of pillar is "r" from one end of the slab

So here net torque must be balance with respect to pillar to be in balanced state

So here we will have

$Mg(r - L/2) = mg(L/2 - 8)$