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Vladimir79 [104]
3 years ago
12

What forces are acting on a dropped book that falls to the floor?

Physics
1 answer:
pav-90 [236]3 years ago
5 0

Answer:

i believe its gravity

Explanation:

when something falls it is gravity doing its job

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A ball is dropped and falls with an acceleration of 9.8 m/s2 downward. It hits the ground with a velocity of 49 m/s downward. Ho
il63 [147K]
Hey!

NOTE-:

u= initial velocity
v= final velocity
g= acceleration due to gravity
t= time

u= 0
v= 49 m/s
t=?
g= 9.8 m/s^2

Using first equation of motion -

v-u=at
49-0= 9.8×t
49 = 9.8t
49/9.8= t
t= 5 second


Hope it helps...!!!
6 0
3 years ago
the two forces f1 and f2 acting at A have a resultant force of Fr= -100lb. determine the magnitude and coordinate direction angl
faust18 [17]
Two forces F<span>1 and </span>F<span>2 act on the screw eye. The resultant force </span>FR<span> has a magnitude of 125 lb and the coordinate direction angles shown in (Figure 1) . Determine the magnitude of </span>F<span>2. Determine the coordinate direction angle </span>α<span>2 of </span>F<span>2. Determine the coordinate direction angle </span>β<span>2 of </span>F<span>2. Determine the coordinate direction angle </span>γ<span>2 of </span>F<span>2.</span>
8 0
3 years ago
An object is said to move from a position of 10m East to a position of 5m west. Determine the object's distance travelled.
motikmotik

Answer:

5 i think

Explanation:

4 0
3 years ago
A 0.l ‑kilogram block is attached to an initially unstretched spring of force constant k = 40 N/m as shown right. The block is d
GalinKa [24]

Answer:

The maximum potential energy of the system is 0.2 J

Explanation:

Hi there!

When the spring is stretched, it acquires potential energy. When released, the potential energy is converted into kinetic energy. If there is no friction nor any dissipative forces, all the potential energy will be converted into kinetic energy according to the energy conservation theorem.

The equation of elastic potential energy (EPE) is the following:

EPE = 1/2 · k · x²

Where:

k = spring constant.

x = stretching distance.

The elastic potential energy is maximum when the block has no kinetic energy, just before releasing it.

Then:

EPE = 1/2 · 40 N/m · (0.1 m)²

EPE = 0.2 J

The maximum potential energy of the system is 0.2 J

8 0
3 years ago
A length of copper wire carries a current of 14 A, uniformly distributed through its cross section. The wire diameter is 2.5 mm,
gavmur [86]

Answer:

a. ρ_\beta=1.996J/m^3

b. U_E=9.445x10^{-15} J/m^3

Explanation:

a. To find the density of magnetic field given use the gauss law and the equation:

i=14A, d=2.5mm, R=3.3Ω, l=1 km, E_o=8.85x10^{-12}F/m, u_o=4*x10^{-7}H/m

ρ_\beta=\frac{\beta^2}{2*u_o}

ρ_\beta=\frac{1}{2*u_o}*(\frac{u_o*i^2}{2\pi *r})^2

ρ_\beta=\frac{u_o*i^2}{8\pi*r}=\frac{4\pi *10^{-7}H/m*(14A)^2}{8\pi*(1.25x10^{-3}m)^2}

ρ_\beta=1.996J/m^3

b. The electric field can be find using the equation:

U_E=\frac{1}{2}*E_o*E^2

E=(\frac{i*R}{l})^2

U_E=\frac{1}{2}*8.85x10^{-12}*(\frac{14A*3.3}{1000m})^2

U_E=9.445x10^{-15} J/m^3

4 0
3 years ago
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