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2 years ago

If an object moves with constant acceleration, its velocity must

1 answer:

5 0

Answer:20 m/s. ... 10 m/s. If an object moves with constant acceleration, its velocity must. change by the same amount each second.

Explanation:

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What is the iv and the DV in a rocket

alekssr [168]

an independent variable (IV) that you can change. • Typical IV's might be number of fins, length of rocket, type of material, mass of the rocket. • Choose one dependent variable (DV) and make three launches per DV for each IV for a total of. six launches..

Explanation:

6 0

3 years ago

A 0.100-kg stone rests on a frictionless, horizontal surface. A bullet of mass 6.00 g, travel- ing horizontally at 350 m>s, s

mariarad [96]

Answer:

Magnitude is 25.8 m/s and direction is $35.5^{o}$

Explanation:

From the law of conservation of linear momentum

$m_{b}v_{ib}+ m_{s}v_{is}= m_{b}v_{fb}+ m_{b}v_{fs}$ where $m_{b}$ and $m_{s}$ are masses of bullet and stones respectively, $v_{ib}$ and $v_{is}$ are the initial velocities of bullet and stone respectively, $v_{fb}$ and $v_{fs}$ are the final velocities of bullet and stone respectively

Substituting 6g=0.006 Kg for mass of bullet, 0.1Kg for mass of stone, 350 m/s for initial velocity of bullet and 250 m/s as final velocity for stone but initial velocity of stone is zero

$0.006 Kg *350 m/s (\hat i)+0.1 Kg*0=0.006 Kg* 250 m/s (\hat j)+0.1*v_{fs}$

$2.1 Kg.m/s(\hat i)=1.5 Kg.m/s(\hat j)+ 0.1*v_{fs}$

$0.1*v_{fs}=2.1 Kg.m/s(\hat i)- 1.5 Kg.m/s(\hat j)$

$v_{fs}=21 Kg.m/s(\hat i)- 15 Kg.m/s(\hat j)$

$|v_{fs}|=\sqrt {(v_{x}^{2}+v_{y}^{2})}$

Substituting 21 for $v_{x}$ and 15 for $v_{y}$

$|v_{fs}|=\sqrt {(21^{2}+15^{2})}=25.8 m/s$

To find direction

$tan\theta=\frac {v_{y}}{v_{x}}$

$\theta=tan^{-1}(\frac {15}{21})=35.5^{o}$

Therefore, magnitude is 25.8 m/s and direction is $35.5^{o}$

8 0

4 years ago

A 48.0-turn circular coil of radius 5.50 cm can be oriented in any direction in a uniform magnetic field having a magnitude of 0

Andrew [12]

Answer:

Explanation:

Given that,

Number of turn is 48

N=48

Radius is 4.8cm

r=0.048m

Magnetic Field

B=0.48T

Current in coil

i=23.3mA

i=0.233A

Maximum Torque?

Maximum torque occur at angle 90°

Torque is given as

τ = N•I•A•B•sinθ

Where N is number of turn =48

I is current in coil =0.233A

A is area of circular coil form

Area of a circle is given as

A=πr²

A=π×0.048²

A=0.007238m²

B is magnetic field =0.48T

Maximum torque occurs at 90°

τ = N•I•A•B•sinθ

τ=48×0.233×0.007238×0.48×Sin90

τ = 0.0389Nm

This torque is large enough to exert the coil

4 0

4 years ago

Read 2 more answers

The annoying sound from a mosquito is produced when it beats its wings at the average rate of 600. wingbeats per second. What is

jok3333 [9.3K]

The frequency is exactly the rate at which the bug beats its wings.
If that "600" that you mentioned is 600 beats per second, then
THAT's the frequency . . . 600 per second. (600 Hz)

3 0

4 years ago

Learning Goal: To practice Problem-Solving Strategy 7.2 Problems Using Mechanical Energy II. The Great Sandini is a 60.0-kg circ

andreev551 [17]

Answer:

v = 15.8 m/s

Explanation:

Let's analyze the situation a little, we have a compressed spring so it has an elastic energy that will become part kinetic energy and a potential part for the man to get out of the barrel, in addition there is a friction force that they perform work against the movement. So the variation of mechanical energy is equal to the work of the fictional force

$W_{fr}$ = ΔEm = $Em_{f}$ -Em₀

Let's write the mechanical energy at each point

Initial

Em₀ = Ke = ½ k x²

Final

$Em_{f}$ = K + U = ½ m v² + mg y

Let's use Hooke's law to find compression

F = - k x

x = -F / k

x = 4400/1100

x = - 4 m

Let's write the energy equation

fr d = ½ m v² + mgy - ½ k x²

Let's clear the speed

v² = (fr d + ½ kx² - mg y) 2 / m

v² = (40 4.00 + ½ 1100 4² - 60.0 9.8 2.50) 2/60.0

v² = (160 + 8800 - 1470) / 30

v = √ (229.66)

v = 15.8 m/s

5 0

4 years ago

Read 2 more answers