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Anton [14]
3 years ago
7

9. An unbalanced force of 400 N is applied to a 10 kg object. What is the acceleration of the object?

Physics
1 answer:
scoray [572]3 years ago
5 0
F=ma. Therefore
a = f/m
400/10 = 40 ms^-2
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All of the following are examples of positive direction except:
charle [14.2K]

Answer:

right

Explanation:

think of a compass theres N.E.S.W north, east, south, west

8 0
3 years ago
Felecia left her home to visit her daughter, driving 45mph. Her husband waited for the dog sitter to arrive and left home 20 min
andre [41]

Felicia** drove at 45 mph.  After 1/3 of an hour, when her husband started out, she was (45/3) = 15 miles from home.

Husband drove at 55 mph.  He gained (55-45) = 10 miles on Felicia each hour.  That is, he closed 10 miles of the gap each hour.

It took him (15/10) = 1.5 hours to catch up to her.  

That's <em>1hr 30min after HE started out</em>, and <em>1hr 50min after SHE left the house</em>.

** I've spelled the wife's name as 'Felicia' throughout my solution.  I'm pretty sure there's a better chance of this being the correct spelling rather than 'Felecia', and I never trust anybody who can't spell their own name.

7 0
3 years ago
Read 2 more answers
A charge of +0.08 C moves to the right due to a 4 N force exerted by an electric field. What is the magnitude and direction of t
Otrada [13]

This is late but the answer is 50 n/c right

4 0
3 years ago
Read 2 more answers
An account voltage is connected to an RLC series circuit of resistance 5ohms, inductance 3mH, and a capacitor of 0.05f. calculat
Nesterboy [21]

Answer:

12.99 Hz

Explanation:

Resonance is said to occur in an RLC circuit when maximum current is obtained from the circuit. Hence at resonance; XL=XC and Z=R.

XL= inductive reactance, XC= capacitive reactance, Z= impedance, R= resistance

The resonance frequency is given by;

fo= 1/2π√LC

L= 3×10^-3 H

C= 0.05 F

π= 3.142

Substituting values;

fo= 1/2×3.142√3×10^-3 × 0.05

fo= 12.99 Hz

Therefore the resonance frequency of the RLC circuit is 12.99 Hz

8 0
3 years ago
A rifle bullet with mass 8.00 g and initial horizontal velocity 280 m/s strikes and embeds itself in a block with mass 0.992 kg
anyanavicka [17]

Answer:

0.4113772 s

Explanation:

Given the following :

Mass of bullet (m1) = 8g = 0.008kg

Initial horizontal Velocity (u1) = 280m/s

Mass of block (m2) = 0.992kg

Maxumum distance (x) = 15cm = 0.15m

Recall;

Period (T) = 2π√(m/k)

According to the law of conservation of momentum : (inelastic Collison)

m1 * u1 = (m1 + m2) * v

Where v is the final Velocity of the colliding bodies

0.008 * 280 = (0.008 + 0.992) * v

2.24 = 1 * v

v = 2.24m/s

K. E = P. E

K. E = 0.5mv^2

P.E = 0.5kx^2

0.5(0.992 + 0.008)*2.24^2 = 0.5*k*(0.15)^2

0.5*1*5.0176 = 0.5*k*0.0225

2.5088 = 0.01125k

k = 2.5088 / 0.01125

k = 223.00444 N/m

Therefore,

Period (T) = 2π√(m/k)

T = 2π√(0.992+0.008) / 233.0444

T = 2π√0.0042910

T = 2π * 0.0655059

T = 0.4113772 s

6 0
3 years ago
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