answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
stira [4]
2 years ago
15

A rocket starts from rest and moves upward from the surface of the earth. for the first 10.0 s of its motion, the vertical accel

eration of the rocket is given by ay=(2.60m/s3)t, where the +y-direction is upward.
Physics
1 answer:
Vikki [24]2 years ago
5 0

acceleration of rocket is given here as

a_y = 2.60* t

now we know that

\frac{dv}{dt} = 2.60t

now integrating both sides

\int dv = \int 2.60t dt

v = 2.60\frac{t^2}{2}

v = 1.30 t^2

here since its given that rocket will accelerate for t = 10 s

so here we have

v = 1.30 * 10^2

v = 130 m/s

so after t = 10 s the speed of rocket will be 130 m/s upwards

You might be interested in
A potential difference of 10.0 volts exists between two points, A and B, within an electric field. What is the
Viefleur [7K]

Answer:

1. 5.0 x 10^2 C

Explanation:

V=W/Q

10 = 2.0 x 10^-2/Q

Q = 2.0 x 10^-2/ 10

Q = 5.0 x 10^2 C

7 0
2 years ago
What magnitude charge creates a 1.0 n/c electric field at a point 1.0 m away?
Stolb23 [73]

Answer:

1.1\cdot 10^{-10}C

Explanation:

The electric field produced by a single point charge is given by:

E=k\frac{q}{r^2}

where

k is the Coulomb's constant

q is the charge

r is the distance from the charge

In this problem, we have

E = 1.0 N/C (magnitude of the electric field)

r = 1.0 m (distance from the charge)

Solving the equation for q, we find the charge:

q=\frac{Er^2}{k}=\frac{(1.0 N/c)(1.0 m)^2}{9\cdot 10^9 Nm^2c^{-2}}=1.1\cdot 10^{-10}C

8 0
2 years ago
A 2 kg stone is tied to a 0.5 m string and swung around a circle at a constant angular velocity of 12 rad/s. the angular momentu
Illusion [34]
Starting from the angular velocity, we can calculate the tangential velocity of the stone:
v=\omega r= (12 rad/s)(0.5 m)= 6 m/s

Then we can calculate the angular momentum of the stone about the center of the circle, given by
L=mvr
where
m is the stone mass
v its tangential velocity
r is the radius of the circle, that corresponds to the length of the string.

Substituting the data of the problem, we find
L=(2 kg)(6 m/s)(0.5 m)=6 kg m^2 s^{-1}
3 0
2 years ago
A Porsche 944 Turbo has a rated engine power of 217 hp. 30% of the power is lost in the drive train, and 70% reaches the wheels.
shutvik [7]

Answer:

a = 6.53 m/s^2

v = 11.5689 m/s

Explanation:

Given data:

engine power is 217 hp

70 % power reached to wheel

total mass ( car + driver) is 1530 kg

from the data given

2/3 rd of weight is over the wheel

w = 2/3rd mg

maximum force

F = \mu W

we know that F = ma

ma =  \mu (2/3 mg)

a_{max} = 2/3(1.00) (9.8) = 6.53 m/s^2

the new power is p  = 70\% P_[max} = 0.7 P_{max}

P =f_{max} v

0.7P_{max} = ma_{max} v

solving for speed v

v =0.7 \times \frac{P_{max}}{ma_{max}}

v = 0.7 \frac{217 [\frac{746 w}{1 hp}]}{1500 \times 6.53}

v = 11.5689 m/s

7 0
2 years ago
A proton initially moves left to right along the x-axis at a speed of 2.00 x 103 m/s. It moves into an uniform electric field, w
FinnZ [79.3K]

Answer:

E = 1.04*10⁻¹ N/C

Explanation:

Assuming no other forces acting on the proton than the electric field, as this is uniform, we can calculate the acceleration of the proton, with the following kinematic equation:

vf^{2} -vo^{2} = 2*a*x

As the proton is coming at rest after travelling 0.200 m to the right,  vf = 0, and x = 0.200 m.

Replacing this values in the equation above, we can solve for a, as follows:

a = \frac{vo^{2}*mp}{2*x} = \frac{(2.00e3m/s)^{2}}{2*0.2m} = 1e7 m/s2

According to Newton´s 2nd Law, and applying the definition of an electric field, we can say the following:

F = mp*a = q*E

For a proton, we have the following values:

mp = 1.67*10⁻²⁷ kg

q = e = 1.6*10⁻¹⁹ C

So, we can solve for E (in magnitude) , as follows:

E = \frac{mp*a}{e} =\frac{1.67e-27kg*1e7m/s2}{1.6e-19C} = 1.04e-1 N/C

⇒ E = 1.04*10⁻¹ N/C

5 0
2 years ago
Other questions:
  • In the Atwood machine shown below, m1 = 2.00 kg and m2 = 6.05 kg. The masses of the pulley and string are negligible by comparis
    6·1 answer
  • A 75-hp (shaft output) motor that has an efficiency of 91.0 percent is worn out and is to be replaced by a high- efficiency moto
    5·1 answer
  • Write a hypothesis about how the height of the cylinder affects the temperature of the water. Use the "if . . . then . . . becau
    14·2 answers
  • In springboard diving, the diver strides out to the end of the board, takes a jump onto its end, and uses the resultant spring-l
    15·2 answers
  • An aluminum rod and a nickel rod are both 5.00 m long at 20.0 degree Celsius. The temperature of each is raised to 70.0 degrees
    7·1 answer
  • 1. A 930-kg car traveling 56 km/h comes to a complete stop in 2.0 s. What is the
    13·1 answer
  • A 1.5 m cylinder of radius 1.1 cm is made of a complicated mixture materials. Its resistivity depends on the distance x from the
    9·1 answer
  • When the sun’s rays are at an angle of 39°, the distance from the top of Dakota’s head to the tip of her shadow is 77 inches. Ab
    14·1 answer
  • The surface is tilted to an angle of 37 degrees from the horizontal, as shown above in Figure 3. The blocks are each given a pus
    7·1 answer
  • You are a particle physicist at the Large Hadron Collider who is tasked with designing an apparatus to separate annihilation pro
    10·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!