1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
Alisiya [41]
4 years ago
15

A roofer drops a nail that hits the ground traveling at 26 m/s. How fast was the nail traveling 1 second before it hits the grou

nd?
Physics
1 answer:
ELEN [110]4 years ago
5 0
This problem can be solved using a kinematic equation. For this case, the following equation is useful:

v_final = v_initial + at

where,
v_final = final velocity of the nail
v_initial = initial velocity of the nail
a = acceleration due to gravity = 9.8 m/s^2
t = time 

First, we determine the time it takes for the nail to hit the ground. We know that the initial velocity is 0 m/s since the nail was only dropped. It has a final velocity of 26 m/s. We substitute these values to the equation and solve for t:

26 = 0 + 9.8*t
t = 26/9.8 = 2.6531 s

The problem asks the velocity of the nail at t = 1 second. We then subtract 1 second from the total time 2.6531 with v_final as unknown.

v_final = 0 + 9.8(2.6531-1) = 16.2004 m/s.

Thus, the nail was traveling at a speed of 16. 2004 m/s, 1 second before it hit the ground. 

You might be interested in
What is the “p” process for elements heavier than iron
tia_tia [17]
Supernova nucleosynthesis is also thought to be responsible for the creation of rarerelements heavier than iron<span> and nickel, in the last few seconds of a type II supernova event.</span>
3 0
3 years ago
Two flat 4.0 cm × 4.0 cm electrodes carrying equal but opposite charges are spaced 2.0 mm apart with their midpoints opposite ea
serious [3.7K]

Answer:

1.77 x 10^-8 C

Explanation:

Let the surface charge density of each of the plate is σ.

A = 4 x 4 = 16 cm^2 = 16 x 10^-4 m^2

d = 2 mm

E = 2.5 x 10^6 N/C

ε0 = 8.85 × 10-12 C2/N ∙ m2

Electric filed between the plates (two oppositively charged)

E = σ / ε0

σ = ε0 x E

σ = 8.85 x 10^-12 x 2.5 x 10^6 = 22.125 x 10^-6 C/m^2

The surface charge density of each plate is ± σ / 2

So, the surface charge density on each = ± 22.125 x 10^-6 / 2

                                                                 = ± 11.0625 x 10^-6 C/m^2  

Charge on each plate = Surface charge density on each plate x area of each plate

Charge on each plate = ± 11.0625 x 10^-6  x 16 x 10^-4 = ± 1.77 x 10^-8 C

7 0
3 years ago
if you drop a softball from just above your knee, the kinetic energy of the ball just before it hits the ground is about 1 jule,
ipn [44]
False.

The mass of a softball is approximately 200 g (0.2 kg), while the knees are located approximately at 30 cm (0.3 m) from the ground. It means that the gravitational potential energy of the ball when it is dropped is
U=mgh=(0.2 kg)(9.81 m/s^2)(0.3 m)=0.6 J

This corresponds to the total mechanical energy of the ball at the moment it is dropped, because there is no kinetic energy (the ball starts from rest). Then the ball is dropped, and just before it hits the ground, all this energy is converted into kinetic energy: but energy cannot be created, so its final kinetic energy cannot be greater than 0.6 J.
7 0
3 years ago
If the specific surface energy for magnesium oxide is 1.0 J/m2 and its modulus of elasticity is (225 GPa), compute the critical
alexdok [17]

Answer:

The critical stress required for the propagation of an initial crack              \sigma_{c} =  21.84 M pa

Explanation:

Given data

Modulus of elasticity E = 225 × 10^{9} \frac{N}{m^{2} }

Specific surface energy for magnesium oxide is \gamma_{s} = 1 \frac{J}{m^{2} }

Crack length (a) = 0.3 mm = 0.0003 m

Critical stress is given by \sigma_{c}^{2} } = \frac{2 E \gamma}{\pi a} -------- (1)

⇒ 2 E \gamma_{s} = 2 × 225 × 10^{9} × 1 = 450 × 10^{9}

⇒ \pi a = 3.14 × 0.0003 = 0.000942  

⇒ Put these values in equation 1 we get

⇒ \sigma_{c}^{2} } = \frac{450  }{0.000942} 10^{9}

⇒ \sigma_{c}^{2} } = 4.77 × 10^{14}

⇒ \sigma_{c} = 2.184 × 10^{7} \frac{N}{m^{2} }

⇒ \sigma_{c} =  21.84 \frac{N}{mm^{2} }

⇒ \sigma_{c} =  21.84 M pa

This is the critical stress required for the propagation of an initial crack.

4 0
3 years ago
A circular-motion addict of mass 83.0 kg rides a Ferris wheel around in a vertical circle of radius 13.0 m at a constant speed o
Svetllana [295]

Answer:

(A) Time period T = 6.28 SEC

(B) At highest point fore is - 575.83 N

(B) At lowest point force is 1050.97 N      

Explanation:

We have given that mass m = 83 kg

Radius r = 13 m

Speed v = 6.10 m/sec

(A) Time period of the motion is given by T=\frac{2\pi r}{v}=\frac{2\times 3.14\times 13}{6.10}=6.28sec

(b) Net force is given by F_{NET}=\frac{mv^2}{r}=\frac{83\times 6.10^2}{13}=237.571N

Force due to gravity F_{gravity}=mg=83\times 9.8=813.4N

At highest point F_{NORMAL}=F_{NET}-F_{GRAVITY}=237.57-813.4=-575.83

(B) At lowest point F_{NORMAL}=F_{NET}+F_{GRAVITY}=237.57+813.4=1050.97N

5 0
3 years ago
Other questions:
  • Which of the following have the same density. number 26
    10·1 answer
  • A block with mass M attached to a horizontal spring with force constant k is moving with simple harmonic motion having amplitude
    6·1 answer
  • The train 'A' travelled a distance is f 120 km in 3 hours , Whereas another train 'B' travelled a distance of 180 km in 4 hours
    15·2 answers
  • The first PCS technology used a form of time division multiplexing called ____.​ a. ​Time Division Multiple Access (TDMA) b. ​Gl
    10·1 answer
  • Write the letters of the correct answers on the lines at left.
    14·1 answer
  • What is the kinetic energy of a 2000 kg car traveling at a speed of 30 m/s (?65 mph)?
    6·1 answer
  • Why the weight of a body decreases with increases in distance from the earth's surface​
    13·2 answers
  • An object was dropped from the plane with 1000 km above the ground with a mass of 300 kg. Find the acceleration of the object ea
    14·1 answer
  • Explain the Law of Reflection
    9·1 answer
  • A sound wave travels at 379 m/sec and has a wavelength of 8 meters. Calculate its frequency and period.
    13·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!