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stich3 [128]
3 years ago
6

Suppose we replace the original launcher with one that fires the ball upward at twice the speed. We make no other changes. How f

ar behind the cart will the ball land, compared to the distance in the original experiment?
(A) half as far
(B)twice as far
(C) four times as far
(D) the same distance
(E) by a factor not listed above
Physics
1 answer:
expeople1 [14]3 years ago
5 0

Answer:

(C) four times as far

Explanation:

As we know that the range of the launcher is given as

R = \frac{v^2sin(2\theta)}{g}

here we know that all the parameters will remain same but only change is the velocity is doubled

So we will have

R' = \frac{(2v)^2 sin2\theta}{g}

so we have

R' = 4 R

so here correct answer will be

(C) four times as far

You might be interested in
a high speed train travels with an average speed of 250 km/h. the train travels for 2 hrs. how far does the train travel
steposvetlana [31]

Answer:

<h3>The answer is 500 km </h3>

Explanation:

The distance covered by an object given it's velocity and time taken can be found by using the formula

<h3>distance = average velocity × time</h3>

From the question

average speed = 250 km/h

time = 2 hrs

We have

distance = 250 × 2

We have the final answer as

<h3>500 km</h3>

Hope this helps you

5 0
3 years ago
Read 2 more answers
which has a higher acceleration:a 10kg object acted upon with a net force of 20N or an 18kg object acted on by a net force of 20
MA_775_DIABLO [31]
<span>Answer: The acceleration of 10 kg object is greater than that of 18 kg object.

Explanation:
According to Newton's Second law:
F = ma --- (A)

Let's find the acceleration for both 10 kg and 18 kg objects!
The net force on both of these masses = F = 20N

(1) Acceleration of 10 kg object
Mass = m = 10 kg
Plug in the values in equation (A):
20 = 10 * a
Acceleration = a = 2 m/s^2

(2) Acceleration of 18 kg object
Mass = m = 18 kg
Plug in the values in equation (A):

20 = 18 * a
Acceleration = a = 1.11 m/s^2


2 > 1.11; therefore, 10 kg object has the higher acceleration compared to the acceleration of the 18 kg object.</span>
7 0
3 years ago
Read 2 more answers
19,792,000,000 in scientific notation will have how many significant figures
xz_007 [3.2K]

Answer:

= 1.9792 × 10^10

Significant Figures= 5

Explanation:

Look at the attachment below

Hope this helps (:

8 0
3 years ago
The magnetic field in the region between the poles of an electromagnet is uniform at any time, (1 point) but is increasing at th
olga nikolaevna [1]

Answer:

B)

The magnitude of induced emf in the conducting loop is 0.99 mV.

Explanation:

Rate of increase in magnetic field per unit time = 0.090 T/s

Area of the conducting loop = 110 cm^2 = 0.0110 m^2

Electromagnetic induction is the production of an emf or voltage in a coil of wire due to a changing magnetic field through the coil.

Induced e.m.f is given as:

EMF = (-N*change in magnetic field/time)*Area

EMF = rate of change of magnetic field per unit time * Area

EMF = 0.090 * 0.0110

EMF = 0.00099 V

EMF = 0.99 mV

5 0
3 years ago
: A 70 kg man and a 12 kg sled are on the frictionless ice of a frozen lake, 25 m apart but connected by a rope of negligible ma
e-lub [12.9K]

Answer:

x_1 = 3.74m

Explanation:

given,

mass of man = 70 kg

mass of sled = 12 kg

F = m a_s

a_s = \dfrac{F}{m}

a_s = \dfrac{8.2}{12}

a_s = 0.68\ m/s^2

F = m a_m

a_m = \dfrac{F}{m}

a_m = \dfrac{8.2}{70}

a_m = 0.12\ m/s^2

x_1+x_2 = 25

\dfrac{1}{2}a_ct^2+ \dfrac{1}{2}a_mt^2 = 25

(a_c+a_m)t^2=50

(0.12+0.68)t^2=50

t = \sqrt{\dfrac{50}{0.8}}

t = 7.90 s

x_1 = \dfrac{1}{2}a_ct^2

x_1 = 0.5\times 0.12 \times 7.90^2

x_1 = 3.74m

5 0
3 years ago
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