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3 years ago

11

You have gathered information about which websites your classmates visit to research science questions. What type of graph would

best represent this information?
A. pie chart
B. bar graph
C. line graph
D. scatter plot

1 answer:

swat323 years ago

8 0

A bar graph would be best so B. Bar graph.

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A pitcher throws a 0.144-kg baseball toward the batter so that it crosses home plate horizontally and has a speed of 42 m/s just

Solnce55 [7]

(a) 12.8 kg m/s

The impulse delivered by the bat on the baseball is equal to the change in momentum of the baseball:

$I=\Delta p = m(v-u)$

where we have

m = 0.144 kg is the mass of the ball

v = -47 m/s is the final velocity of the ball

u = 42 m/s is the initial velocity of the ball

Substituting into the equation, we find

$I=(0.144 kg)(-47 m/s-(42 m/s))=-12.8 kg m/s$

And since we are interested in the magnitude only,

$I=12.8 kg m/s$

(b) 2.78 kN

The impulse exerted on the ball is also equal to the product between the average force and the contact time:

$I=F\Delta t$

where

F is the average force exerted on the ball

$\Delta t=0.0046 s$ is the contact time

Solving the formula for F, we find

$F=\frac{I}{\Delta t}=\frac{12.8 kg m/s}{0.0046 s}=2783 N = 2.78 kN$

(c) The force exerted on the ball is much larger (1988 times more) than the weigth of the ball

The weight of the ball is given by

$W=mg$

where

m = 0.144 kg is the mass of the ball

g = 9.8 m/s^2 is the acceleration due to gravity

Solving the equation for W, we find

$W=(0.144 kg)(9.8 m/s^2)=1.4 N$

So as we see, the force exerted on the ball (2783 N) is almost 2000 times larger than the weight of the ball (1.4 N):

$\frac{F}{W}=\frac{2783 N}{1.4 N}=1988$

8 0

3 years ago

Find the acceleration of a body whose velocity increases from 11ms-1 to 33ms-1 in 10 seconds

Anvisha [2.4K]

Acceleration is equivalent to the rate of change of velocity.

Calculate the difference of velocity, and divide by the total time it took for the change to happen.

$(33 \frac{m}{s} - 11 \frac{m}{s} ) \div 10seconds \\ = 2 \frac{m}{ {s}^{2} }$
That is the answer.

8 0

4 years ago

Read 2 more answers

A very loud train whistle has an acoustic power output of 100 W. If the sound energy spreads out spherically, what is the intens

valentina_108 [34]

Answer

given,

Power output of loud train = 100 W

Distance of intensity level from train= 100 m

I₀ = 10⁻¹² W/m²

Intensity of sound = ?

we know,

$I = \dfrac{P}{Area}$

$I = \dfrac{P}{4\pi r^2}$

$I = \dfrac{100}{4\pi 100^2}$

$I =7.957\times 10^{-4}\ W/m^2$

Intensity level

= $10 log(\dfrac{I}{I_0})$

= $10 log(\dfrac{7.975\times 10^{-4}}{10^{-12}})$

= $10 \times 8.900$

= 89 dB

7 0

3 years ago

A 50kg boy runs at a speed of 10.0m/s and jumpsonto a cart

andriy [413]

Answer:

the mass of the cart is 150 kg

Explanation:

given,

mass of boy(m) = 50 kg

speed of boy (v)= 10 m/s

initial velocity of cart (u) = 0

final velocity of cart(V) = 2.5 m/s

mass of the cart(M) = ?

m v + M u = (m + M ) V......................(1)

50× 10 + 0 = (50 + M ) 2.5

M = $\dfrac{500}{2.5} - 50$

M = 150 Kg

hence, the mass of the cart is 150 kg

4 0

3 years ago

Read 2 more answers

A 0.6 kg ball is initially at rest on a frictionless, horizontal surface. It is struck by a 0.4 kg ball initially moving with a

Dafna1 [17]

Answer:

140

Explanation:

7 0

3 years ago