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katovenus [111]
4 years ago
7

What is the displacement of this time velocity graph i need help asap

Physics
1 answer:
maw [93]4 years ago
3 0

Answer:

b is the answer

Explanation:

because i did this before

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Austin invested $11,000 in an account paying an interest rate of 5.7% compounded quarterly. Assuming no deposits or withdrawals
daser333 [38]

Answer:

15448

Explanation:

Compounded Quarterly:

A=P\left(1+\frac{r}{n}\right)^{nt}

A=P(1+

n

r

​

)

nt

Compound interest formula

P=11000\hspace{35px}r=0.057\hspace{35px}t=6\hspace{35px}n=4

P=11000r=0.057t=6n=4

Given values

A=11000\left(1+\frac{0.057}{4}\right)^{4(6)}

A=11000(1+

4

0.057

​

)

4(6)

Plug in values

A=11000(1.01425)^{24}

A=11000(1.01425)

24

Simplify

A=15448.0290759

A=15448.0290759

Use calculator

3 0
3 years ago
Read 2 more answers
Who invented gravity
PilotLPTM [1.2K]

Answer: joe

Explanation:

5 0
3 years ago
Read 2 more answers
if 130N centripetal force is needed to keep a 0.45kg ball that is attached to a string that is 0.7m long to complete 5 full rota
andrezito [222]

Answer:

centripetal acceleration of the ball is 6.9 m/s/s

tangential speed of the ball is 2.2 m/s

Explanation:

As we know that ball complete 5 rotations in 10 seconds

so frequency of rotation of ball is given as

f = \frac{5}{10} = 0.5 Hz

now we know that angular frequency is given as

\omega = 2\pi f

\omega = 2(\pi)(0.5)

\omega = \pi rad/s

Now centripetal acceleration is given as

a_c = \omega^2 R

a_c = \pi^2 (0.7)

a_c = 6.9 m/s^2

now the velocity of the ball at this angular frequency is given as

v = R\omega

v = 0.7 (\pi)

v = 2.2 m/s

4 0
3 years ago
Determine the launch speed of a horizontally launched projectile that lands 26.3m from the base of a 19.3m high cliff.
atroni [7]

The launch velocity of the projectile is 13.28 m/s.

What is projectile motion?

The motion of an object thrown in the air under the force of gravity is known as projectile motion.

Since the object is launched horizontally, its initial velocity along the vertical direction is zero. From the second kinematic equation,

s=u*t+(1/2)at^2.

where s is the displacement, t is the time, u is the initial velocity and a is the acceleration. Since the height is decreasing, so it will be taken negative.

For the vertical motion, s=-19.3 m, a=-9.8 m/s^2 and u=0. Put the values in the above equation and solve it.

-19.3 = (0)*t+(1/2)*(-9.8)*t^2

19.3 = (1/2)*(9.8)*t^2

t=1.98 s

Since the velocity along the horizontal direction is constant, the displacement along the horizontal direction is given by the formula,

X=vt

where X is the horizontal displacement, v is the initial horizontal velocity and t is the time.

For the horizontal motion, X=26.3 m and t=1.98 s. Put the values in this equation and solve it.

26.3=v*(1.98)

v=13.28 m/s

The launch velocity is equal to the initial horizontal velocity, so it is equal to 13.28 m/s.

Learn more about projectile motion.

brainly.com/question/11049671

#SPJ4

4 0
2 years ago
A fisherman is fishing from a bridge and is using a "42.0-N test line." In other words, the line will sustain a maximum force of
lara31 [8.8K]

Answer:

(a) 42 N

(b)36.7 N

Explanation:

Nomenclature

F= force test line (N)

W : fish weight  (N)

Problem development

(a) Calculating of weight of the heaviest fish that can be pulled up vertically, when the line is reeled in at constant speed

We apply Newton's first law of equlibrio because the system moves at constant speed:

∑Fy =0

F-W= 0

42N -W =0  

W = 42N

(b) Calculating of weight of the heaviest fish that can be pulled up vertically, when the line is reeled with an acceleration whose magnitude is 1.41 m/s²

We apply Newton's second law because the system moves at constant acceleration:

 m= W/g , m= W/9.8 ,  m:fish mass , W: fish weight g:acceleration due to gravity

∑Fy =m*a

m= W/g , m= W/9.8 ,  m:fish mass , W: fish weight g:acceleration due to gravity

F-W= ( W/9.8 )*a

42-W=  ( W/9.8 )*1.41

42= W+0.1439W

42=1.1439W

W= 42/1.1439

W= 36.7  N

8 0
3 years ago
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