Which of the following forces does NOT act on an object sliding down an incline plane?

If two automobiles have the same velocity, do they have the same acceleration?

gulaghasi [49]

The automobiles do have the same acceleration as the velocity the answer is yes ?

5 0

3 years ago

A cannon is fired from a castle wall at some unknown height above the ground. The cannonball leaves the cannon with speed 30.0m/

Sauron [17]

Answer:

Part a)

$t = 3.85 s$

Part b)

$h = 72.67 m$

Part C)

$v_x = 25.98 m/s$

$v_y = 0$

Part d)

In horizontal direction velocity will remain constant

$v_x = 30 cos30 = 25.98 m/s$

in vertical direction we have

$v_y = -22.77 m/s$

Explanation:

Part a)

Horizontal speed of the cannon

$v = 30.0 m/s$

angle of projection

$\theta = 30^o$

now we have

horizontal speed = $v_x = vcos30 = 30 cos30 =25.98 m/s$

vertical speed = $v_y = vsin30 = 30 sin30 = 15 m/s$

now the time taken by it to cover the distance 100 m from the wall

$x = v_x t$

$100 = 25.98 t$

$t = 3.85 s$

Part b)

Since it hits the ground in the same time

so the height of the castle is given as

$h = \frac{1}{2}gt^2$

$h = \frac{1}{2}(9.81)(3.85^2)$

$h = 72.67 m$

Part C)

At highest point of the projection

the vertical component of the velocity will become zero

so we will have

$v_x = 25.98 m/s$

$v_y = 0$

Part d)

In horizontal direction velocity will remain constant

so we have

$v_x = 30 cos30 = 25.98 m/s$

in vertical direction we have

$v_y = v_i + at$

$v_y = 15 - 9.81(3.85)$

$v_y = -22.77 m/s$

Part e)

8 0

3 years ago

Read 2 more answers

An RLC circuit has a resistance of 200 Ω and an inductance of 15 mH. Its oscillation frequency is 7000 Hz. At time t = 0, the cu

MArishka [77]

Answer: 2.13 × 10^-4 A

Explanation:

Given that the RLC circuit has a resistance R = 200 Ω and an inductance L = 15 mH.

Its oscillation frequency F = 7000 Hz

The initial current I = 25 mA = 25/1000 or 25 × 10^-3 A

Since there is no charge on the capacitor, the current after complete 5 circle will be achieved by using the formula in the attached file.

Please find the attached file for the remaining explanation for the solution.

8 0

3 years ago

B. The coefficient of friction between the tires and the road is 0.850 and the mass of the car is

Mrac [35]

Answer:

156.26N

Explanation:

The data needed are incomplete. Let the acceleration of the body be 3.5m/s²

Other given parameters

Mass = 1.35×10^1 = 13.5kg

coefficient of friction between the tires and the road = 0.850

Acceleration due to gravity = 9.8m/s²

According to Newton's second law:

Fnet = ma

Fnet = Fapp - Ff

Fapp is the applied force

Ff is the frictional force = umg

The equation becomes:

Fapp - Ff = ma

Fapp-umg = ma

Fapp - 0.85(13.5)(9.8) = 13.5(3.5)

Fapp - 109.0125 = 47.25

Fapp = 47.25+109.0125

Fapp = 156.2625N

Hence the applied force that caused the acceleration is 156.26N

Note that the acceleration of the car was assumed. Any value of acceleration can be used for the calculation.

8 0

3 years ago

What is the charge delivered by a current with an average of 0.54 A over 28.3 minutes?

marishachu [46]

Answer:

916.9C

Explanation:

Using Q= It

Q= 0.54A x 1698s

= 916.9C

6 0

4 years ago