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a concave mirror has a focal length of 18 cm. where will an image form if an object is placed 58 cm from the mirror

MAXImum [283]

Answer:

here

Explanation:

8 0

3 years ago

Which graph accurately shows the relationship between kinetic energy and speed as speed increases?

mixer [17]

Answer:

B

Explanation:

kinetic energy (KE) is the energy possessed by moving bodies. It can be expressed as:

KE = $\frac{1}{2}$ m $v^{2}$

Where: m is the mass of the object, and v its speed.

For example, a stone of mass 2kg was thrown and moves with a speed of 3 m/s. Determine the kinetic energy of the stone.

Thus,

KE = $\frac{1}{2}$ x 2 x $3^{2}$

= 9

KE = 9.0 Joules

Assume that the speed of the stone was 4 m/s, then its KE would be:

KE = $\frac{1}{2}$ x 2 x $4^{2}$

= 16

KE = 16.0 Joules

Therefore, it can be observed that as speed increases, the kinetic energy increases. Thus option B is appropriate.

3 0

3 years ago

A rocket initially at rest accelerates at a rate of 99. 0 meters/second2. Calculate the distance covered by the rocket if it att

creativ13 [48]

The rocket will cover $1 \times 10^3\rm \ m$ distance in 4. 5 s. Acceleration can be defined as the change in velocity.

<h3>What is acceleration?</h3>

Acceleration can be defined as the change in speed or the direction of the object.

From kinamatic equation:

$D = v_{t} +\dfrac 12at^2$

Where,

$D$ - final velocity = 445 m/s

$v_0$ - initial valocity = 0 m/s

$a$ - acceleration = 99. 0 m/s²

$t$ - time = 4. 50 s

Put the values in the formula,

$D = 0\times ( 4.5) + \dfrac12\times (99)(4.5)^2\\\\D = 1002 {\rm \ m}\\\\D = 1 \times 10^3\rm \ m$

Therefore, the rocket will cover $1 \times 10^3\rm \ m$ distance in 4. 5 s.

Learn more about Acceleration :

brainly.com/question/2697545

7 0

3 years ago

A boy and his younger sister are at the zoo on a hot day. They each buy a cold lemonade. The boy buys a large lemonade, and his

FromTheMoon [43]

Answer: A and C (i took the test)

Explanation: Hope this helps:)

7 0

4 years ago

A particle of charge 2.0 x 10^-8C experiences an upward force of magnitude 4.0 x10^-6 when it is placed in a particular point in

koban [17]

Answer:

a) The electric field at that point is $2.0\times 10^{2}$ newtons per coulomb.

b) The electric force is $2.0\times 10^{-6}$ newtons.

Explanation:

a) Let suppose that electric field is uniform, then the following electric field can be applied:

$E = \frac{F_{e}}{q}$ (1)

Where:

$E$ - Electric field, measured in newtons per coulomb.

$F_{e}$ - Electric force, measured in newtons.

$q$ - Electric charge, measured in coulombs.

If we know that $F_{e} = 4.0\times 10^{-6}\,N$ and $q = 2.0\times 10^{-8}\,C$ , then the electric field at that point is:

$E = \frac{4.0\times 10^{-6}\,N}{2.0\times 10^{-8}\,C}$

$E = 2.0\times 10^{2}\,\frac{N}{C}$

The electric field at that point is $2.0\times 10^{2}$ newtons per coulomb.

b) If we know that $E = 2.0\times 10^{2}\,\frac{N}{C}$ and $q = 1.0\times 10^{-8}\,C$ , then the electric force is:

$F_{e} = E\cdot q$

$F_{e} = \left(2.0\times 10^{2}\,\frac{N}{C} \right)\cdot (1.0\times 10^{-8}\,C)$

$F_{e} = 2.0\times 10^{-6}\,N$

The electric force is $2.0\times 10^{-6}$ newtons.

7 0

3 years ago