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2 years ago

Please help! The image produced by a concave mirror is ? .

Physics

2 answers:

Alexeev081 [22]2 years ago

6 0

Answer:

is a reflection.

The image is real light rays actually focus at the image location). As the object moves towards the mirror the image location moves further away from the mirror and the image size grows (but the image is still inverted). When the object is that the focal point, the image is at infinity.

Explanation:

SSSSS [86.1K]2 years ago

3 0

Answer:

virtual or real

Explanation:

You might be interested in

A battery is connected to a 10 resistor and produces a current of 0.2 A in the circuit. If the resistor is replaced with a 20 re

drek231 [11]

Answer:

0.1 A

Explanation:

From the question,

V = IR............ Equation 1

Where V = Voltage, I = current, R = Resistance.

Given: I = 0.2 A, R = 10 ohms.

Substitute into equation 1

V = 0.2(10)

V = 2 volt,

If the resistor is replaced with a 20 resistor, The nwe current is

I = V/R................ Equation 2

I = 2/20

I = 0.1 A

4 0

2 years ago

Spacetime interval: What is the interval between two events if in some given inertial reference frame the events are separated b

shepuryov [24]

Answer:

a. $\Delta s ^2 = 8.0888 \ 10^{17} m^2$

b. $\Delta s ^2 = 3.0234 \ 10^{16} m^2$

c. $\Delta s ^2 = 3.0234 \ 10^{20} m^2$

Explanation:

The spacetime interval $\Delta s^2$ is given by

$\Delta s ^2 = \Delta (c t) ^ 2 - \Delta \vec{x}^2$

please, be aware this is the definition for the signature ( + - - - ), for the signature (- + + + ) the spacetime interval is given by:

$\Delta s ^2 = - \Delta (c t) ^ 2 + \Delta \vec{x}^2$ .

Lets work with the signature ( + - - - ), and, if needed in the other signature, we can multiply our interval by -1.

$\Delta \vec{x}^2 = (7.5 \ 10 \ m)^2$

$\Delta \vec{x}^2 = 5,625 m^2$

$\Delta (c t) ^ 2 = (299,792,458 \frac{m}{s} \ 3 \ s)^2$

$\Delta (c t) ^ 2 = (899,377,374 \ m)^2$

$\Delta (c t) ^ 2 = 8.0888 \ 10^{17} m^2$

$\Delta s ^2 = 8.0888 \ 10^{17} m^2 - 5,625 m^2$

$\Delta s ^2 = 8.0888 \ 10^{17} m^2$

$\Delta \vec{x}^2 = (5 \ 10 \ m)^2$

$\Delta \vec{x}^2 = 2,500 m^2$

$\Delta (c t) ^ 2 = (299,792,458 \frac{m}{s} \ 0.58 \ s)^2$

$\Delta (c t) ^ 2 = (173,879,625.6 \ m)^2$

$\Delta (c t) ^ 2 = 3.0234 \ 10^{16} m^2$

$\Delta s ^2 = 3.0234 \ 10^{16} m^2 - 2,500 m^2$

$\Delta s ^2 = 3.0234 \ 10^{16} m^2$

$\Delta \vec{x}^2 = (5 \ 10 \ m)^2$

$\Delta \vec{x}^2 = 2,500 m^2$

$\Delta (c t) ^ 2 = (299,792,458 \frac{m}{s} \ 58 \ s)^2$

$\Delta (c t) ^ 2 = (1.73879 \ 10^{10} \ m)^2$

$\Delta (c t) ^ 2 = 3.0234 \ 10^{20} m^2$

$\Delta s ^2 = 3.0234 \ 10^{20} m^2 - 2,500 m^2$

$\Delta s ^2 = 3.0234 \ 10^{20} m^2$

5 0

2 years ago

When a high-energy proton or pion traveling near the speed of light collides with a nucleus, it may travel 2.5 x 10-15 m before

Tasya [4]

Explanation:

It is given that,

When a high-energy proton or pion traveling near the speed of light collides with a nucleus, $d=2.5\times 10^{-15}\ m$

Speed of light, $c=3\times 10^{8}\ m\s$

Let t is the time interval required for the strong interaction to occur. The speed is given by :

$c=\dfrac{d}{t}$

$t=\dfrac{d}{c}$

$t=\dfrac{2.5\times 10^{-15}\ m}{3\times 10^{8}\ m/s}$

$t=8.33\times 10^{-24}\ s$

So, the time interval required for the strong interaction to occur is $8.33\times 10^{-24}\ s$ . Hence, this is the required solution.

8 0

3 years ago

Why is friction like applied force but different from gravity?

kirill115 [55]

Answer:

Friction is when a force is applied or done by weight dragging onto something.

Explanation:

Gravity is when an object is getting pulled toward the center of what is attracting it. And applied force is when someone/sommething is applying force.

3 0

3 years ago

Read 2 more answers

An object weighs 32 newtons. What is its mass if a gravitometer indicates that g = 8.25 m/s?

Elan Coil [88]

Explanation:

weight=mass×g

32=mass×8.25

mass=

$\frac{128}{33} = 3.878$ kg

8 0

2 years ago