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lianna [129]
2 years ago
15

A car moving at a speed of 20 m/s, then accelerates uniformly at until it reaches a speed of What distance does it travel during

the time it is accelerating?
a)50 m
B)54 m
C)58 m
D)62 m
E)66 m
Physics
1 answer:
Andru [333]2 years ago
6 0
E because you have to drive
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What forces affect a toy car as it moves down a ramp?
Amanda [17]

Answer:

gravity

Explanation:

3 0
2 years ago
Read 2 more answers
(1) A net force of 265 N accelerates a bike and rider at 2.30 m/s. What is the mass of the
kati45 [8]

Answer: 115.2kg

Explanation:

Net force = 265 N

Acceleration of bike & rider = 2.30m/s2 (The SI unit of acceleration is m/s2)

Mass of the bike and rider together = ?

Since force is the product of the mass of an object and the acceleration by which it moves, Force = Mass x Acceleration

265N = Mass x 2.30m/s2

Mass = (265N/2.30m/s2)

Mass = 115.2 kg

Thus, the Mass of the bike and rider together is 115.2kg

6 0
3 years ago
Can someone please help
ki77a [65]

Answer:

Acceleration of that planet is 30 \frac{m}{s^{2} }.

Given:

initial speed of hammer = 0 \frac{m}{s}

time = 1 s

distance = 15 m

To find:

Acceleration due to gravity = ?

Formula used:

Distance covered by hammer is given by,

s = ut + \frac{1}{2} a t^{2}

s = distance

u = initial speed of hammer

t = time taken by hammer to reach ground

a = acceleration

Solution:

Distance covered by hammer is given by,

s = ut + \frac{1}{2} a t^{2}

s = distance

u = initial speed of hammer

t = time taken by hammer to reach ground

a = acceleration

u = 0

t = 1 s

s = 15 m

a = g

Thus substituting these value in above equation.

15 = 0 + \frac{1}{2} g 1^{2}

g = 15 × 2

g = 30 \frac{m}{s^{2} }

Thus, acceleration of that planet is 30 \frac{m}{s^{2} }.

8 0
3 years ago
lasie4. A 4 kg object is displaced to the right by a distance of 12 m underthe influence of the following forces: a 17 Nforce pu
Oksanka [162]

The work done by a constant force in a rectilinear motion is given by:

W=Fd\cos\theta

where F is the magnitude of the force, d is the distance and θ is the angle between the force and the displacement vector.

In this case we have two forces then we need to add the work done by each of them; for the first force we have a magnitude of 17 N, a displacement of 12 m and and angle of 0° (since both the displacement and the force point right); for the second force we have a magnitude of 36 N, a displacement of 12 m and an angle of 30°. Plugging these values we have that the total work is:

\begin{gathered} W=(17)(12)\cos0+(36)(12)\cos30 \\ W=578.123 \end{gathered}

Therefore, the total work done is 578.123 J and the answer is option E

6 0
10 months ago
Show all work please I am stuck​
MissTica

Answer:

Explanation:

1 meter = 39.37 inches

meters x inches/meters = inches

6.23x10^-4 m x 39.37 in./m=0.245...=2.45x10^-2

5 0
2 years ago
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