Date:

4. Two forces act on a 2 kg object as shown. What is the magnitude of the acceleration of the object?

aleksandrvk [35]

The resultant force on the object is

∑ F = 〈0, 8〉 N + 〈6, 0〉 N = 〈6, 8〉 N

which has a magnitude of

F = √((6 N)² + (8 N)²) = √(100 N²) = 10 N

By Newton's second law, the acceleration has magnitude a such that

F = m a

10 N = (2 kg) a

a = (10 N) / (2 kg)

a = 5 m/s²

so the answer is B.

4 0

2 years ago

A 1500 W electric heater is plugged into the outlet of a 120 V circuit that has a 20 A circuit breaker. You plug an electric hai

Mila [183]

Answer:

Explanation:

Current drawn by electric heater = power/volt =1500/120 = 12.5 A.

current drawn by hair drier at 600 watt = 600/120 =5 A

current drawn by hair drier at 900 watt = 900/120 = 7.5 A.

Total current drawn by heater and hair drier used at 900 watt

= 12.5 + 7.5 = 20 A

Breaking current =20 A

So fuse will trip at this point .

3 0

2 years ago

A capacitor is charged until it holds 5.0 j of energy. it is then connected across a 10-kω resistor. in 13.6 ms , the resistor d

prohojiy [21]

Answer:

The capacite is C=5.32 uF using the equations of voltage and energy in capacitance

Explanation:

The energy holds is 5 J and the resistor dissipates 2J so the energy total is 3J

Using:

$V_{t}= V_{o}e^{\frac{-t}{R*C} }$

Voltage in this case is the energy dissipated so

$E_{t}= E_{o}e^{\frac{-t}{R*C} }$

$\frac{\sqrt{E_t} }{\sqrt{E_o} } = e^{\frac{-t}{R*C} }$

$\frac{\sqrt{3 J} }{\sqrt{5J} } = e^{\frac{-13.6ms}{10kw*C} }$

Using the equation to find capacitance

$ln 0.775= e^{\frac{-13.6 x10^{3} }{10x10^{3}*C }} \\ln(0.775)= ln * e^{\frac{-13.6 x10^{3s} }{10x10^{3}*C }} \\\\ln(0.775)= {\frac{-13.6 x10^{3} }{10x10^{3}*C }} \\C= \frac{-13.6 x10^{-3} }{10x10x^{3}*ln(0.775) }$

$C= 5.32x10^{-6}$ F

C= 5.32 uF because u is the symbol for micro that is equal to $10^{-6}$

8 0

3 years ago

PLEASE HURRYYY

babymother [125]

your answer is make up artist

4 0

3 years ago

6000 kg train moving 5 m/sec to east collides with 5000 kg train moving 3 m/sec to west. What is their velocity

Svetlanka [38]

The final velocity is 1.37 m/s east

Explanation:

We can solve this problem by using the law of conservation of momentum: in fact, in absence of external forces, the total momentum of the two trains must be conserved before and after the collision.

So we can write:

$p_i = p_f\\m_1 u_1 + m_2 u_2 = (m_1+m_2)v$