All categories

3 years ago

Acceleration changes and objects...

Physics

1 answer:

brilliants [131]3 years ago

3 0

Answer:

speed and direction

Explanation:

Acceleration is the rate of change of speed and direction.

You might be interested in

Student bikes to school by traveling first dN = 1.10 miles north, then dW = 0.300 miles west, and finally dS = 0.200 miles south

melisa1 [442]

Taking the vertical component of the displacement
1.1 - 0.2 = 0.9 mile
The horizontal component of the displacement
-0.3 mile

The magnitude of the displacement is
√[ (0.9)² + (-0.3) ] = 0.95 mile

The direction is
θ = tan-1 (-0.3/0.9)
θ = 161.57 degrees.

8 0

3 years ago

Where must an object be placed to form an image 30.0 cm from a diverging lens with a focal length of 43.0 cm?

Schach [20]

Using lens equation;

1/o + 1/i = 1/f; where o = Object distance, i = image distance (normally negative), f = focal length (normally negative)

Substituting;

1/o + 1/-30 = 1/-43 => 1/o = -1/43 + 1/30 = 0.01 => o = 1/0.01 = 99.23 cm

Therefore, the object should be place 99.23 cm from the lens.

6 0

3 years ago

An airplane is flying at 635 km per hour at an altitude of 35,000 m. What is its velocity?

Elden [556K]

Distance 350 Km

Time 1 hour

Velocity = 350 : 1 =

350Km/h

your answer is a

5 0

3 years ago

The radius of curvature of a spherical mirror is 20cm.What is its focal length?

hichkok12 [17]

Answer:

$\boxed{\sf Focal \ length = 10 \ cm}$

Given:

Radius of curvature (R) of a spherical mirror = 20

To Find:

Focal length (f)

Explanation:

Formula:

$\boxed{ \bold{\sf Focal \ length \ (f) = \frac{Radius \ of \ curvature \ (R)}{2}}}$

Substituting value of R in the equation:

$\sf \implies f = \frac{20}{2}$

$\sf \implies f = \frac{ \cancel{2} \times 10}{ \cancel{2}}$

$\sf \implies f = 10 \: cm$

5 0

3 years ago

Read 2 more answers

Hans Langseth's beard measured 5.33 m in 1927. Consider two charges, q1 = 2.42 nC and an unspecified charge, q2, are separated 5

schepotkina [342]

Answer:

-7.89 * 10^(-9) C

Explanation:

Parameters given:

q1 = 2.42 nC = 2.42 * 10^(-9) C

Distance between q1 and q2 = 5.33 m

q3 = 1.0 nC = 1 * 10^(-9) C

Distance between q1 and q3 = 1.9 m

Distance between q2 and q3 = 5.33 - 1.9 = 3.43 m

The net force acting on q3 is:

F = F(q1, q3) + F(q2, q3)

F = (k*q1*q3)/1.9² + (k*q2*q3)/3.43²

F = (9 * 10^(9) * 2.42 * 10^(-9) * 1 * 10^(-9))/3.61 + (9 * 10^(9) * q2 * 1 * 10^(-9))/11.7649

F = 6.033 * 10^(-9) + 0.765*q2

If the net force is zero:

0 = 6.033 * 10^(-9) + 0.765*q2

-0.765*q2 = 6.033 * 10^(-9)

=> q2 = -[6.033 * 10^(-9)]/0.765

q2 = -7.89 * 10^(-9) C

3 0

3 years ago