If the net force acting on an object is 0 N, you can be sure thr forces acting on the object are,

When an object is at a distance of twice the focal length from a concave lens, the image produced is virtual and smaller than th

Elenna [48]

The right answer for the question that is being asked and shown above is that: "The image produced is virtual and of the same size as the object." the image if the object is shifted closer to the lens to a point one focal length away from it is that The image produced is virtual and of the same size as the object.<span>
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3 0

4 years ago

Read 2 more answers

Jada is rowing a boat across a river that has a current of 5 m/s in the ˆ j direction. Leanne, standing on the shore, observes J

olchik [2.2K]

Answer: d. 8.25 m/s

Explanation:

We are given that Current= 5 m/s in j direction

Velocity= 8 m/s i + 3 m/s j

Now, we have to find Jada's speed with respect to the water.

First we find Jada's velocity with respect to water

v= (8 i + 3 j) - (5 j)

v= 8i - 2 j

To find the speed, we take the magnitude of this velocity vector we have

|v|= $\sqrt{(8)^2+(-2)^2}$

|v|= $\sqrt{68}$ = 8.246 m/s

which comes out to be around = 8.25 m/s

So option d is correct.

5 0

3 years ago

4. How much time does it take for a student running at a speed of 5 m/s to cover a distance of 2,000 m?

Monica [59]

Answer:

6 Minutes 40 Seconds or 400 Seconds

Explanation:

Time to cover a distance of 5m = 1 Second

Time to cover a distance of 2000m = 2000÷5

= 400 Seconds

After converting 400 Seconds into minutes it will become 6 minutes 40 seconds.

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6 0

3 years ago

The four-wheel-drive all-terrain vehicle has a mass of 385 kg with center of mass G2. The driver has a mass of 75 kg with center

jenyasd209 [6]

The coefficient of friction is missing and it has a value of μ = 0.4

Answer:

a = 3.924 m/s²

Explanation:

I've attached the kinematic free body diagram.

Taking the sum of all upward and downward forces,

EFy = 0;

N1 + N2 - m_p•g - m_v•g = 0

N1 + N2 = m_p•g + m_v•g

Where;

N1 and N2 are the normal reactions at the wheels

m_p is the mass of the driver

m_v is the mass of the vehicle

g is the acceleration due to gravity.

Plugging in the relevant values in the question,we obtain;

N1 + N2 = (385 + 75) x 9.81

N1 + N2 = 4512.6N - - - (eq1)

Now, taking sum of all horizontal forces;

EFx = (m_p + m_v) x a

So,

μ(N1 + N2) = (mp + mv) x a

Thus,

0.4(N1 + N2) = (385 + 75)a

0.4(N1 + N2) = 460a

N1 + N2 = 1150a

From eq(1),N1 + N2 = 4512.6N

Thus,

1150a = 4512.6N

a = 4512.6/1150

a = 3.924 m/s²

Therefore, the acceleration, a = 3.924 m/s²

7 0

3 years ago

An alternative to CFL bulbs and incandescent bulbs are light-emitting diode (LED) bulbs. A 16-W LED bulb can replace a 100-W inc

Yanka [14]

Answer:

LED bulb = 0.145 A

Incandescent bulb = 0.909 A

CFL bulb = 0.218 A

Explanation:

Given:

Power rating of LED bulb (P₁) = 16 W

Power rating of incandescent bulb (P₂) = 100 W

Power rating of CFL bulb (P₃) = 24 W

Terminal voltage across the circuit (V) = 110 V

We know that, power is related to terminal voltage and current drawn as:

$P=VI$

Express this in terms of 'I'. This gives,

$I=\frac{P}{V}$

Now, calculate the current drawn in each bulb using their respective values.

For LED bulb, $P_1=16\ W, V=110\ V$

So, current drawn is given as:

$I_1=\frac{16\ W}{110\ V}=0.145\ A$

For incandescent bulb, $P_2=100\ W, V=110\ V$

So, current drawn is given as:

$I_2=\frac{100\ W}{110\ V}=0.909\ A$

For CFL bulb, $P_3=24\ W, V=110\ V$

So, current drawn is given as:

$I_3=\frac{24\ W}{110\ V}=0.218\ A$

Therefore, the currents drawn through LED bulb, incandescent bulb and CFL bulb are 0.145 A, 0.909 A and 0.218 A respectively.

5 0

4 years ago