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Mariana [72]
3 years ago
10

A 12-kg object is moving rightward with a constant velocity of 4 m/s. How much net force is required to keep the object moving w

ith
the same speed and in the same direction?
Physics
1 answer:
Nadusha1986 [10]3 years ago
3 0
C an s and yes ihavetotypemore
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What force causes weight? How is weight affected as an object moves farther away from the center of the earth?
stich3 [128]

Gravity affects weight of an object

Its weight reduces as it moves away from the center as gravity is strongest near the core and reduces as you move away

Hope this helps C:


3 0
4 years ago
Read 2 more answers
The specific heat of a certain type of metal is 0.128 J/(g⋅∘C).0.128 J/(g⋅∘C). What is the final temperature if 305 J305 J of he
Makovka662 [10]

Answer:

45.3°C

Explanation:

Heat gained = mass × specific heat × increase in temperature

q = mC (T − T₀)

Given C = 0.128 J/g/°C, m = 94.0 g, q = 305 J, and T₀ = 20.0°C:

305 J = (94.0 g) (0.128 J/g/°C) (T − 20.0°C)

T = 45.3°C

6 0
3 years ago
An X-ray tube emits X-rays with a wavelength of 1.0 x 10-11 m. Calculate the potential that must be applied across the X-ray tub
sergey [27]

Answer:

1.24 x 10 to the 5 ev = 124,000 ev          its B

Explanation:

E = hc/lambda = 1.24 ev-micrometer/1.0x10 to the -5 micrometers = 1.24 x 10 to the 5 ev = 124,000 ev

h = Planck's constant = 6.626 × 10 to the -34 joule·s

c = speed of light = 2.998 × 10 to the 8 m/s

lambda is the given wavelength

E is the desired photon energy

3 0
3 years ago
Read 2 more answers
Two rings of radius 5 cm are 20 cm apart and concentric with a common horizontal x-axis. The ring on the left carries a uniforml
Yanka [14]

Answer:

The electric field due to the right ring at a location midway between the two rings is 2.41\times10^{3}\ V/m

Explanation:

Given that,

Radius of first ring = 5 cm

Radius of second ring = 20 cm

Charge on the left of the ring = +30 nC

Charge on the right of the ring = -30 nC

We need to calculate the electric field due to the right ring at a location midway between the two rings

Using formula of  electric field

E=\dfrac{1}{4\pi\epsilon_{0}}\times\dfrac{qx}{(x^2+R^2)^{\frac{3}{2}}}

Put the value into the formula

E=\dfrac{9\times10^{9}\times30\times10^{-9}\times0.1}{((0.1)^2+(0.2)^2)^{\frac{3}{2}}}

E=2.41\times10^{3}\ V/m

Hence, The electric field due to the right ring at a location midway between the two rings is 2.41\times10^{3}\ V/m

3 0
4 years ago
A conducting sphere contains positive charge distributed uniformly over its surface. Which statements about the potential due to
stira [4]

Answer:

a. FALSE

b. FALSE

c. TRUTH

d. FALSE

e. FALSE

Explanation:

To determine which statements are truth or false you focus in the following formula, for the electric potential generated by a conducting sphere:

V=\frac{Q}{4\pi \epsilon_o R}      inside the sphere

V'=\frac{Q}{4\pi \epsilon_o r}      for r > R (outside the sphere)

R: radius of the sphere

ε0: dielectric permittivity of vacuum

Q: charge of the sphere

As you can notice, inside the sphere the potential is constant. Inside the sphere, the potential is the same. Outside the surface the potential decreases as 1/r, being r the distance to the center of the sphere.

Hence, you can conclude:

a. The potential at the center of the sphere is zero. FALSE

b.The potential is lowest, but not zero, at the center of the sphere. FALSE

c. The potential at the center of the sphere is the same as the potential at the surface. TRUTH

d. The potential at the center is the same as the potential at infinity. FALSE

e. The potential at the surface is higher than the potential at the center. FALSE

8 0
3 years ago
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