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3 years ago

Describe the 3 types of projectiles.

2 answers:

7 0

Answer:Although any object in motion through space (for example a thrown baseball, kicked football, fired bullet, thrown arrow, stone released from catapult) are projectiles, they are commonly found in warfare and sports.

Explanation:

bulgar [2K]3 years ago

5 0

<u>ANS</u>

Three types of projectiles —

The round ball, The bullet, and shot—

used in muzzleloaders.

<u>Th</u><u>ank</u> <u>You</u> !!!!!

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Which item(s) would be sufficient to make a circuit?

masya89 [10]

Explanation :

A circuit is the representation of the path of the flow of current. The circuit can be either closed or open.

When the switch is off the circuit is closed circuit and when the switch is not connected the circuit is open.

The items that are sufficient to make a circuit are as follows :

Voltage source like a battery.
Resistors or electrical equipment like heater, motor etc.

Other components can be ammeter, voltmeter, ac source, variable resistors etc.

7 0

3 years ago

Read 2 more answers

A 49 kg bear slides, from rest, 11 m down a lodgepole pine tree, moving with a speed of 3.3 m/s just before hitting the ground.

hichkok12 [17]

Answer:

a) $\Delta U_g=-5.3kJ$

b) $K=0.27kJ$

c) $F_f=0.45kN$

Explanation:

the gravitational potential energy is given by:

$U_g=m.g.h\\$

$\Delta U_g=m.g.h_f-m.g.h_i\\\Delta U_g=49kg*9.8m/s^2*(0m-11m)\\\Delta U_g=-5.3kJ$

The kinetic energy is given by:

$K=\frac{1}{2}m.v^2\\$

the initial kinetic energy is zero because the motion started from rest, so:

$K=\frac{1}{2}*49kg*(3.3m/s^2)^2\\K=0.27kJ$

applying the conservation of energy theorem:

$U_g-W_f=K_f\\W_f=-(\Delta K+\Delta U)\\W_F=5.3kJ-0.27kJ\\W_F=-5.0kJ$

The work done by the friction force is given by:

$W_f=F_f.h.cos(\theta)\\$

the angle of the force is 180 degrees because it's against the movement:

$F_f=\frac{W_f}{h.cos(\theta)}\\\\F_f=\frac{-5.0kJ}{11m.cos(180^o)}\\\\F_f=0.45kN$

8 0

4 years ago

A car of mass 750 kg accelerates away from traffic lights. At the end of the first 100 m it has reached a speed

malfutka [58]

the work done on the car by the force of its engine is 78,000 J.

" The work done on the car by the force of friction is 24,000 J.

Increasing the car's kinetic energy at the end of the first 100 m is 54,000J

a. Completed work = force x distance. Engine output = 780 N, that is,

780 N x 100 m = 78,000 J.

b. Completed work = force x distance. Friction force = 240 N, that is,

240 N x 100 m = 24,000 J.

c. Kinetic energy = 1 \ 2 x m x v2

= 1 \ 2 x 750 kg x 12 squared = 375 x 144 = 54,000 J.

<h3>How powerful is the engine of a car? </h3>

Mainstream car and truck engines typically produce 100-400 pounds. -Torque feet. This torque is generated by the engine piston as it moves up and down on the engine crankshaft, causing the engine to rotate (or twist) continuously.

Learn more about work done here: brainly.com/question/25573309

#SPJ10

5 0

2 years ago

a child's toy consists of a piece of plastic attached to a spring. the spring is compressed against the floor a distance of 2.0

Nat2105 [25]

Answer:

1.7N

Explanation:

Force = kx

Where x = spring compression and

K = spring constant

K =85N/m

x = 2.0cm / 100

= 0.02m

Force = 85 x 0.02

= 1.7N

5 0

3 years ago

What is the difference in PE

Blababa [14]

Answer:

831.4 J

Explanation:

Info given:

mass adult: 82.5kg

mass child: 14.7kg

height of wall: 1.25m

g = 9.81m/s^2

PE = mgh

For adult:

mgh = (82.5kg)(9.81m/s^2)(1.25m) = 1011.65 J

For child:

mgh = (14.7kg)(9.81m/s^2)(1.25m) = 180.25 J

Difference in PE: 1011.65 J - 180.25 J = 831.4 J

3 0

3 years ago